Math Problem
Thread Starter
Joined: Jun 2001
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From: UK
Math Problem
What number when you multipy the sums of its digits by 8 takes you back to the original number?
I know the answer to this (clever clogs aint I?!) but can anyone prove mathematically that there are no other numbers that fit this criteria (0bviously Zero doesn't count!).
I know the answer to this (clever clogs aint I?!) but can anyone prove mathematically that there are no other numbers that fit this criteria (0bviously Zero doesn't count!).
Jet Blast Rat
Joined: Jan 2001
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From: Sarfend-on-Sea
So you have a series of figures n1 to ni, all single digits, such that n1 + 10n2 +100n3 + ... = 8 x (n1 + n2 + n3 + ...)
= 8 x n1 + 8 x n2 + 8 x n3 + ...
Seeing it like this there is certainly an upper cap, as the multiplication by 8 is eclipsed by the multiplication by 100 of the figure in the third column, and by 1000 of the figure in the 4th column. There must also be a lower limit above 9, as the figure in the first column is a unit, yet multiplied by 8 on the other side of the equation.
Certainly more than 20, to account for the 8s from the 10s column and then some from the units (since 10 itself does not fit except in octal). Certainly less than 100 because all hundreds only give 8 each as any digit does on the other side of the equation, and the maximum units figure is 9, giving 72 which cannot make up for the rest even at 100. Any figure in the 10s column of course only make things worse, giving 10 on the side that is already higher and only 8 again on the other side.
The only options are therefore 24, 32, 40, 48, 56, 64, 72, 80, 88 and 96, the muliples of 8 in this range. The answer is of course 72.
= 8 x n1 + 8 x n2 + 8 x n3 + ...
Seeing it like this there is certainly an upper cap, as the multiplication by 8 is eclipsed by the multiplication by 100 of the figure in the third column, and by 1000 of the figure in the 4th column. There must also be a lower limit above 9, as the figure in the first column is a unit, yet multiplied by 8 on the other side of the equation.
Certainly more than 20, to account for the 8s from the 10s column and then some from the units (since 10 itself does not fit except in octal). Certainly less than 100 because all hundreds only give 8 each as any digit does on the other side of the equation, and the maximum units figure is 9, giving 72 which cannot make up for the rest even at 100. Any figure in the 10s column of course only make things worse, giving 10 on the side that is already higher and only 8 again on the other side.
The only options are therefore 24, 32, 40, 48, 56, 64, 72, 80, 88 and 96, the muliples of 8 in this range. The answer is of course 72.
I'matightbastard
Joined: Jul 2001
Posts: 1,747
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From: Texas
Or, let the answer (n) be a two digit number with the first digit represented by a and the second digit represented by b
n = 10a + b
We are told that the sum of the digits multiplied by 8 is the same as the original number
n = 8(a + b)
Therefore,
10a + b = 8(a + b)
10a + b = 8a + 8b
2a = 7b
We know a and b are single digit integers, so the number of solutions to this equation is limited:
Specifically b must be an even number for a to remain an integer
Furthermore b must be <= 2 for a to remain a single digit integer
Consequently b = 2 and a = 7
Therefore the answer is 72
I just can't be arsed to extend the solution to more than two digits, in fact, on reflection, I don't really know why I put this much effort into it
n = 10a + b
We are told that the sum of the digits multiplied by 8 is the same as the original number
n = 8(a + b)
Therefore,
10a + b = 8(a + b)
10a + b = 8a + 8b
2a = 7b
We know a and b are single digit integers, so the number of solutions to this equation is limited:
Specifically b must be an even number for a to remain an integer
Furthermore b must be <= 2 for a to remain a single digit integer
Consequently b = 2 and a = 7
Therefore the answer is 72
I just can't be arsed to extend the solution to more than two digits, in fact, on reflection, I don't really know why I put this much effort into it
