“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
48.7k870156
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
48.7k870156
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
48.7k870156
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
48.7k870156
asked Dec 14 '18 at 11:00
user619818user619818
13418
edited Dec 14 '18 at 11:17
Blue
48.7k870156
asked Dec 14 '18 at 11:00
user619818user619818
13418
edited Dec 14 '18 at 11:17
Blue
48.7k870156
edited Dec 14 '18 at 11:17
Blue
48.7k870156
edited Dec 14 '18 at 11:17
Blue
48.7k870156
48.7k870156
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
473114
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
$endgroup$
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
226k15157275
$endgroup$
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
42.9k356135
$endgroup$
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
20.6k2740
$endgroup$
add a comment |
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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
473114
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
$endgroup$
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
473114
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
$endgroup$
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
473114
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
$endgroup$
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
473114
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
edited Dec 14 '18 at 12:01
AryanSonwatikar
473114
edited Dec 14 '18 at 12:01
AryanSonwatikar
473114
edited Dec 14 '18 at 12:01
AryanSonwatikar
473114
473114
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
answered Dec 14 '18 at 11:39
Angel MorenoAngel Moreno
39715
39715
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
226k15157275
$endgroup$
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
226k15157275
$endgroup$
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
226k15157275
$endgroup$
$$10^na+underbrace{b}_{text{ has } n(>1)text{ digits}}=57biff56b=10^n a$$
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
226k15157275
edited Dec 14 '18 at 11:18
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 14 '18 at 11:05
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
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
42.9k356135
$endgroup$
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
42.9k356135
$endgroup$
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
42.9k356135
$endgroup$
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
42.9k356135
answered Dec 14 '18 at 11:11
TonyKTonyK
42.9k356135
answered Dec 14 '18 at 11:11
TonyKTonyK
42.9k356135
answered Dec 14 '18 at 11:11
TonyKTonyK
42.9k356135
42.9k356135
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
20.6k2740
$endgroup$
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
20.6k2740
$endgroup$
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
20.6k2740
$endgroup$
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
20.6k2740
answered Dec 14 '18 at 11:21
farruhotafarruhota
20.6k2740
answered Dec 14 '18 at 11:21
farruhotafarruhota
20.6k2740
answered Dec 14 '18 at 11:21
farruhotafarruhota
20.6k2740
20.6k2740
add a comment |
add a comment |
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