A Problem about Multiples(Use the least possible value)

Is there a multiple of 41 that consists of all 1s? If so, then what is the smallest such (positive) value?

Write " 0 0 " if there is no such number, otherwise write the answer ex. 1111111 1111111 .


The answer is 11111.

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1 solution

Shu Hung Wang
Nov 12, 2018

You can use a guess and check strategy. For example, 111 is not divisible by 41. You can use a guess and check strategy until you get the answer, in this case, 11111.

Note that the answer is not unique. For example, 1111111111 = 41 × 27100271 1111111111 = 41 \times 27100271 . In fact, every integer consisting of 5 k 5k consecutive 1's is divisible by 41 41 . So for the posted answer to be uniquely correct, you may need to specify that you want the least such integer that satisfies the condition.

Brian Charlesworth - 2 years, 6 months ago

Okay! I will fix that. I meant to say the smallest.

Shu Hung Wang - 2 years, 6 months ago

Yeah well we did this already...

Yukan Wu - 2 years, 6 months ago

Yeah, Ken Wu

Shu Hung Wang - 2 years, 6 months ago

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