### Brain Teasers

# 1/29th

Fun: (2.37)
Difficulty: (2.3)
Puzzle ID: #33282

Submitted By: lessthanjake789 Corrected By: DaleGriffin

Submitted By: lessthanjake789 Corrected By: DaleGriffin

Math
Math brain teasers require computations to solve.

What three digit number is 29 times as large as its last two digits? There is only one answer.

### Hint

Some number ending in X, times 29, equals some number ending in X again.### Answer

725. 725/25 = 29.Hide Hint Show Hint Hide Answer Show Answer

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## Comments

yay first comment.

this teaser is okay.

this teaser is okay.

Good but points off for not showing the math.

Y is two digit number.

X is three digit number.

X + Y = 29 * Y

X = 28 * Y

Some might see that the only single digit factors of 28 are 2, 4, and 7, then quickly realize that X = 700... Most will realize that for X to end in zeroes Y must be a multiple of 5, then quickly get to Y = 25.

Y is two digit number.

X is three digit number.

X + Y = 29 * Y

X = 28 * Y

Some might see that the only single digit factors of 28 are 2, 4, and 7, then quickly realize that X = 700... Most will realize that for X to end in zeroes Y must be a multiple of 5, then quickly get to Y = 25.

I might also spell out that X is a number in even hundreds such that X + Y is the three digit number sought.

yeah, i figured as much from your explanation. yeah, i dont think this is a particularly hard teaser, but i also sometimes like being able to just narrow stuff down by logic. I actually got a problem like this on the AMC12, which is this all-important math test for high school seniors in America, and it wasnt jsut getting the answer, but being able to get it quickly so as to move on to the other problems. once the basics are understood (and you laid them out differently, but neatly enough), its really quite easy.

if you know your math, this could be an easy teaser. but for those average ones, this could be quite hard; solving a problem involving 2 variables with just 1 equation?but if you also know your logic, you will arrive at the correct answer.

of course, i know my math and i know my logic, so got it right!

of course, i know my math and i know my logic, so got it right!

What's the problem with guys like "pating"??

He is always critizicing and telling us how easy he did it (but never giving any solution).

...And the worse is that in general he is wrong!... (read his comments at "Reversible Exponents")

He is always critizicing and telling us how easy he did it (but never giving any solution).

...And the worse is that in general he is wrong!... (read his comments at "Reversible Exponents")

Well, here's my solution:

let x = last 2 digits of the 3-digit number

let y = hundreds digit of the 3-digit number

100y + x = 29x

100y = 28x

To find the value of x, 100y must be a 3-digit number that is a multiple of 100. Just substitute any digit (0 to 9) to y and check if 100y is exactly divisible by 28. only y = 7 will satisfy those conditions. From there, you can solve for x = 100(7)/28 = 25. So your 3-digit number is 725.

Sorry if I offended anyone with my comment but it was never my intention. And I'd be glad (maybe not THAT glad) if my solutions be proven wrong (not just on this teaser, but on the others as well). Coz it would only mean 1 thing, that I got to learn something new!

let x = last 2 digits of the 3-digit number

let y = hundreds digit of the 3-digit number

100y + x = 29x

100y = 28x

To find the value of x, 100y must be a 3-digit number that is a multiple of 100. Just substitute any digit (0 to 9) to y and check if 100y is exactly divisible by 28. only y = 7 will satisfy those conditions. From there, you can solve for x = 100(7)/28 = 25. So your 3-digit number is 725.

Sorry if I offended anyone with my comment but it was never my intention. And I'd be glad (maybe not THAT glad) if my solutions be proven wrong (not just on this teaser, but on the others as well). Coz it would only mean 1 thing, that I got to learn something new!

I didn't both with an equation. The last two digits of the number had to be less than 1000/29 and end in either 0 or 5. Immediately 25 suggested itself since 25 repeats as the last two digits in its products every 8th multiple. Didn't even need to check it since (8x+5) * 25 will end in 25 and 29 = (8*3)+5.

I just took it on faith the this was the only solution since the problem stated that.

I just took it on faith the this was the only solution since the problem stated that.

Great teaser!

@javaguru: 25 repeats as the last two digits in every 4th of its multiple.

@javaguru: 25 repeats as the last two digits in every 4th of its multiple.

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