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 Respectively Divisible (Posted on 2013-02-04)
The numbers 160, 161 and 162 form a set of three consecutive integers (in order) that are divisible respectively by 5, 7 and 9.

Find the first set of four consecutive integers (in order) that are respectively divisible by 5, 7, 9, and 11.

How about first set of five consecutive integers (in order) that are respectively divisible by 5, 7, 9, 11 and 13.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 That's all she wrote (spoiler) Comment 2 of 2 |
Part 1)

160 is the first number that equals 0 mod 5, 6 mod 7, and 7 mod 9.
All other such numbers are 160 + a multiple of 5*7*9, which is to say 160 + 315n, where n is an integer.
We want to find the first of these which = 8 mod 11.

160 = 6 mod 11
315 = 7 mod 11
6+(7*5) = 41 = 8 mod 11, so n = 5

The next series starts with 160+315*5 = 1735
It is 1735, 1736, 1737, 1738

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Part II)

1735 is the first number that equals 0 mod 5, 6 mod 7,  7 mod 9. and 8 mod 11
All other such numbers are 1735 + a multiple of 5*7*9*11, which is to say 1735 + 3465n, where n is an integer.
We want to find the first of these which = 9 mod 13.

1735 = 6 mod 13
3465 = 7 mod 13
6+(7*6) = 48 = 9 mod 13, so n = 6

The next series starts with 1735+3465*6 = 22525
It is 22525, 22526, 22527, 22528, and 22529

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Extra Credit

And the first series of 6 numbers also starts with 22525, because 22530 happens to be divisible by 15.

Edited on February 4, 2013, 7:56 pm
 Posted by Steve Herman on 2013-02-04 17:56:21

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