“A number becomes $57$ times smaller when the first digit is deleted.” Explaining the solution.
$begingroup$
Puzzle: Find a number that becomes 57 times smaller when the first digit is deleted.
Answer: 7125
Explanation:
ab...z = 57 * b...z
X = b...z
aX = 57X
X + a0...0
__ _/
/
k
above is X + a with k zeros
aX = a . 10^k + X
I don't understand from the a0...0 part.
If I relate it to something I do know then I can't see how it makes any sense.
eg
x=3
a=7
so:
ax = 21
imagine we don't know that a = 7 but we do know:
ax = 21
using the logic above we say:
ax = x + a0...0
ax = 3 + a0...0
21 = 3 + 70...0
but 3 + 7 with any number of zeros does not get you 21???
and:
aX = a . 10^k + X
aX = 21
a * 10^k + X
7 * 10^k + 3
but 7 multiplied by any number + 3 does not give us 21???
algebra-precalculus discrete-mathematics
edited Dec 14 '18 at 11:17
Blue
47.7k870151
asked Dec 14 '18 at 11:00
user619818user619818
13418
$endgroup$
add a comment |
$begingroup$
Puzzle: Find a number that becomes 57 times smaller when the first digit is deleted.
Answer: 7125
Explanation:
ab...z = 57 * b...z
X = b...z
aX = 57X
X + a0...0
__ _/
/
k
above is X + a with k zeros
aX = a . 10^k + X
I don't understand from the a0...0 part.
If I relate it to something I do know then I can't see how it makes any sense.
eg
x=3
a=7
so:
ax = 21
imagine we don't know that a = 7 but we do know:
ax = 21
using the logic above we say:
ax = x + a0...0
ax = 3 + a0...0
21 = 3 + 70...0
but 3 + 7 with any number of zeros does not get you 21???
and:
aX = a . 10^k + X
aX = 21
a * 10^k + X
7 * 10^k + 3
but 7 multiplied by any number + 3 does not give us 21???
algebra-precalculus discrete-mathematics
edited Dec 14 '18 at 11:17
Blue
47.7k870151
asked Dec 14 '18 at 11:00
user619818user619818
13418
$endgroup$
add a comment |
$begingroup$
Puzzle: Find a number that becomes 57 times smaller when the first digit is deleted.
Answer: 7125
Explanation:
ab...z = 57 * b...z
X = b...z
aX = 57X
X + a0...0
__ _/
/
k
above is X + a with k zeros
aX = a . 10^k + X
I don't understand from the a0...0 part.
If I relate it to something I do know then I can't see how it makes any sense.
eg
x=3
a=7
so:
ax = 21
imagine we don't know that a = 7 but we do know:
ax = 21
using the logic above we say:
ax = x + a0...0
ax = 3 + a0...0
21 = 3 + 70...0
but 3 + 7 with any number of zeros does not get you 21???
and:
aX = a . 10^k + X
aX = 21
a * 10^k + X
7 * 10^k + 3
but 7 multiplied by any number + 3 does not give us 21???
algebra-precalculus discrete-mathematics
edited Dec 14 '18 at 11:17
Blue
47.7k870151
asked Dec 14 '18 at 11:00
user619818user619818
13418
$endgroup$
Puzzle: Find a number that becomes 57 times smaller when the first digit is deleted.
Answer: 7125
Explanation:
ab...z = 57 * b...z
X = b...z
aX = 57X
X + a0...0
__ _/
/
k
above is X + a with k zeros
aX = a . 10^k + X
I don't understand from the a0...0 part.
If I relate it to something I do know then I can't see how it makes any sense.
eg
x=3
a=7
so:
ax = 21
imagine we don't know that a = 7 but we do know:
ax = 21
using the logic above we say:
ax = x + a0...0
ax = 3 + a0...0
21 = 3 + 70...0
but 3 + 7 with any number of zeros does not get you 21???
and:
aX = a . 10^k + X
aX = 21
a * 10^k + X
7 * 10^k + 3
but 7 multiplied by any number + 3 does not give us 21???
algebra-precalculus discrete-mathematics
algebra-precalculus discrete-mathematics
edited Dec 14 '18 at 11:17
Blue
47.7k870151
asked Dec 14 '18 at 11:00
user619818user619818
13418
edited Dec 14 '18 at 11:17
Blue
47.7k870151
asked Dec 14 '18 at 11:00
user619818user619818
13418
edited Dec 14 '18 at 11:17
Blue
47.7k870151
edited Dec 14 '18 at 11:17
Blue
47.7k870151
edited Dec 14 '18 at 11:17
Blue
47.7k870151
47.7k870151
asked Dec 14 '18 at 11:00
user619818user619818
13418
asked Dec 14 '18 at 11:00
user619818user619818
13418
asked Dec 14 '18 at 11:00
user619818user619818
13418
13418
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let : $$ 1<a<9$$ $$ 0 le n $$ $$ 0 < b < 10^n $$
Then the equation to be resolved is: $$ 10^n a + b = 57 b $$
$$ 2^n 5^n a = 2^3 cdot 7 cdot b $$
$$ 2^{n-3} cdot 5^n cdot a = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} cdot a = 7 cdot b $$
If $ a = 7 $ : $$ 5^3 cdot 10^{n-3} cdot 7 = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} = b $$
$$ 125 cdot 10^{n-3} = b $$
$$ a=7, b in {125, 1250, 12500, ...} $$
If $ a neq 7 $(and $a neq 0$ precondition): No solutions, the left part of equation is not divisible by 7 and the right always divisible by 7.
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
$endgroup$
add a comment |
$begingroup$
$$10^na+underbrace{b}_{text{ has } n(>1)text{ digits}}=57biff56b=10^n a$$
Now $dfrac{10^na}7=8b$ which is an integer
As $(7,10^n)=1,7|aimplies a=7$ as $0<a<10$
$implies8b=10^nimplies b=dfrac{10^n}{2^3}=10^{n-3}cdot5^3=5^n2^{n-3}$
As $b$ is an integer, $nge3$
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Here, $aX$ doesn't mean $a$ times $X$; it means "the decimal number with first digit $a$ followed by the digits of $X$". So in your example, $aX$ would be $73$, not $21$.
It's true that there is nothing in the notation itself to tell you this; but it's the only interpretation that makes sense in context. (And to add to the confusion, when they write $57X$, they do mean $57$ times $X$!)
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
$endgroup$
add a comment |
$begingroup$
Let the number be $overline{a_1a_2...a_n}$. The condition is:
$$overline{a_1a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}+overline{a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}=56overline{a_2...a_n} Rightarrow \
overline{a_2...a_n}=frac{a_1cdot 10^{n-1}}{56}=frac{a_1cdot 2^{n-1}cdot 5^{n-1}}{7cdot 2^3} Rightarrow \
a_1=7; 2^{n-4}in mathbb N Rightarrow nge 4.$$
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039218%2fa-number-becomes-57-times-smaller-when-the-first-digit-is-deleted-explainin%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let : $$ 1<a<9$$ $$ 0 le n $$ $$ 0 < b < 10^n $$
Then the equation to be resolved is: $$ 10^n a + b = 57 b $$
$$ 2^n 5^n a = 2^3 cdot 7 cdot b $$
$$ 2^{n-3} cdot 5^n cdot a = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} cdot a = 7 cdot b $$
If $ a = 7 $ : $$ 5^3 cdot 10^{n-3} cdot 7 = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} = b $$
$$ 125 cdot 10^{n-3} = b $$
$$ a=7, b in {125, 1250, 12500, ...} $$
If $ a neq 7 $(and $a neq 0$ precondition): No solutions, the left part of equation is not divisible by 7 and the right always divisible by 7.
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
$endgroup$
add a comment |
$begingroup$
Let : $$ 1<a<9$$ $$ 0 le n $$ $$ 0 < b < 10^n $$
Then the equation to be resolved is: $$ 10^n a + b = 57 b $$
$$ 2^n 5^n a = 2^3 cdot 7 cdot b $$
$$ 2^{n-3} cdot 5^n cdot a = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} cdot a = 7 cdot b $$
If $ a = 7 $ : $$ 5^3 cdot 10^{n-3} cdot 7 = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} = b $$
$$ 125 cdot 10^{n-3} = b $$
$$ a=7, b in {125, 1250, 12500, ...} $$
If $ a neq 7 $(and $a neq 0$ precondition): No solutions, the left part of equation is not divisible by 7 and the right always divisible by 7.
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
$endgroup$
add a comment |
$begingroup$
Let : $$ 1<a<9$$ $$ 0 le n $$ $$ 0 < b < 10^n $$
Then the equation to be resolved is: $$ 10^n a + b = 57 b $$
$$ 2^n 5^n a = 2^3 cdot 7 cdot b $$
$$ 2^{n-3} cdot 5^n cdot a = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} cdot a = 7 cdot b $$
If $ a = 7 $ : $$ 5^3 cdot 10^{n-3} cdot 7 = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} = b $$
$$ 125 cdot 10^{n-3} = b $$
$$ a=7, b in {125, 1250, 12500, ...} $$
If $ a neq 7 $(and $a neq 0$ precondition): No solutions, the left part of equation is not divisible by 7 and the right always divisible by 7.
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
$endgroup$
Let : $$ 1<a<9$$ $$ 0 le n $$ $$ 0 < b < 10^n $$
Then the equation to be resolved is: $$ 10^n a + b = 57 b $$
$$ 2^n 5^n a = 2^3 cdot 7 cdot b $$
$$ 2^{n-3} cdot 5^n cdot a = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} cdot a = 7 cdot b $$
If $ a = 7 $ : $$ 5^3 cdot 10^{n-3} cdot 7 = 7 cdot b $$
$$ 5^3 cdot 10^{n-3} = b $$
$$ 125 cdot 10^{n-3} = b $$
$$ a=7, b in {125, 1250, 12500, ...} $$
If $ a neq 7 $(and $a neq 0$ precondition): No solutions, the left part of equation is not divisible by 7 and the right always divisible by 7.
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
edited Dec 14 '18 at 12:01
AryanSonwatikar
31312
31312
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
37915
37915
add a comment |
add a comment |
$begingroup$
$$10^na+underbrace{b}_{text{ has } n(>1)text{ digits}}=57biff56b=10^n a$$
Now $dfrac{10^na}7=8b$ which is an integer
As $(7,10^n)=1,7|aimplies a=7$ as $0<a<10$
$implies8b=10^nimplies b=dfrac{10^n}{2^3}=10^{n-3}cdot5^3=5^n2^{n-3}$
As $b$ is an integer, $nge3$
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$10^na+underbrace{b}_{text{ has } n(>1)text{ digits}}=57biff56b=10^n a$$
Now $dfrac{10^na}7=8b$ which is an integer
As $(7,10^n)=1,7|aimplies a=7$ as $0<a<10$
$implies8b=10^nimplies b=dfrac{10^n}{2^3}=10^{n-3}cdot5^3=5^n2^{n-3}$
As $b$ is an integer, $nge3$
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$10^na+underbrace{b}_{text{ has } n(>1)text{ digits}}=57biff56b=10^n a$$
Now $dfrac{10^na}7=8b$ which is an integer
As $(7,10^n)=1,7|aimplies a=7$ as $0<a<10$
$implies8b=10^nimplies b=dfrac{10^n}{2^3}=10^{n-3}cdot5^3=5^n2^{n-3}$
As $b$ is an integer, $nge3$
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$$10^na+underbrace{b}_{text{ has } n(>1)text{ digits}}=57biff56b=10^n a$$
Now $dfrac{10^na}7=8b$ which is an integer
As $(7,10^n)=1,7|aimplies a=7$ as $0<a<10$
$implies8b=10^nimplies b=dfrac{10^n}{2^3}=10^{n-3}cdot5^3=5^n2^{n-3}$
As $b$ is an integer, $nge3$
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
edited Dec 14 '18 at 11:18
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$begingroup$
Here, $aX$ doesn't mean $a$ times $X$; it means "the decimal number with first digit $a$ followed by the digits of $X$". So in your example, $aX$ would be $73$, not $21$.
It's true that there is nothing in the notation itself to tell you this; but it's the only interpretation that makes sense in context. (And to add to the confusion, when they write $57X$, they do mean $57$ times $X$!)
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
$endgroup$
add a comment |
$begingroup$
Here, $aX$ doesn't mean $a$ times $X$; it means "the decimal number with first digit $a$ followed by the digits of $X$". So in your example, $aX$ would be $73$, not $21$.
It's true that there is nothing in the notation itself to tell you this; but it's the only interpretation that makes sense in context. (And to add to the confusion, when they write $57X$, they do mean $57$ times $X$!)
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
$endgroup$
add a comment |
$begingroup$
Here, $aX$ doesn't mean $a$ times $X$; it means "the decimal number with first digit $a$ followed by the digits of $X$". So in your example, $aX$ would be $73$, not $21$.
It's true that there is nothing in the notation itself to tell you this; but it's the only interpretation that makes sense in context. (And to add to the confusion, when they write $57X$, they do mean $57$ times $X$!)
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
$endgroup$
Here, $aX$ doesn't mean $a$ times $X$; it means "the decimal number with first digit $a$ followed by the digits of $X$". So in your example, $aX$ would be $73$, not $21$.
It's true that there is nothing in the notation itself to tell you this; but it's the only interpretation that makes sense in context. (And to add to the confusion, when they write $57X$, they do mean $57$ times $X$!)
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
answered Dec 14 '18 at 11:11
TonyKTonyK
42.1k355134
42.1k355134
add a comment |
add a comment |
$begingroup$
Let the number be $overline{a_1a_2...a_n}$. The condition is:
$$overline{a_1a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}+overline{a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}=56overline{a_2...a_n} Rightarrow \
overline{a_2...a_n}=frac{a_1cdot 10^{n-1}}{56}=frac{a_1cdot 2^{n-1}cdot 5^{n-1}}{7cdot 2^3} Rightarrow \
a_1=7; 2^{n-4}in mathbb N Rightarrow nge 4.$$
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
$endgroup$
add a comment |
$begingroup$
Let the number be $overline{a_1a_2...a_n}$. The condition is:
$$overline{a_1a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}+overline{a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}=56overline{a_2...a_n} Rightarrow \
overline{a_2...a_n}=frac{a_1cdot 10^{n-1}}{56}=frac{a_1cdot 2^{n-1}cdot 5^{n-1}}{7cdot 2^3} Rightarrow \
a_1=7; 2^{n-4}in mathbb N Rightarrow nge 4.$$
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
$endgroup$
add a comment |
$begingroup$
Let the number be $overline{a_1a_2...a_n}$. The condition is:
$$overline{a_1a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}+overline{a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}=56overline{a_2...a_n} Rightarrow \
overline{a_2...a_n}=frac{a_1cdot 10^{n-1}}{56}=frac{a_1cdot 2^{n-1}cdot 5^{n-1}}{7cdot 2^3} Rightarrow \
a_1=7; 2^{n-4}in mathbb N Rightarrow nge 4.$$
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
$endgroup$
Let the number be $overline{a_1a_2...a_n}$. The condition is:
$$overline{a_1a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}+overline{a_2...a_n}=57overline{a_2...a_n} Rightarrow \
a_1cdot 10^{n-1}=56overline{a_2...a_n} Rightarrow \
overline{a_2...a_n}=frac{a_1cdot 10^{n-1}}{56}=frac{a_1cdot 2^{n-1}cdot 5^{n-1}}{7cdot 2^3} Rightarrow \
a_1=7; 2^{n-4}in mathbb N Rightarrow nge 4.$$
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
answered Dec 14 '18 at 11:21
farruhotafarruhota
19.7k2738
19.7k2738
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039218%2fa-number-becomes-57-times-smaller-when-the-first-digit-is-deleted-explainin%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
