# Hard Math Word Problem I need help with?

Find a number not greater than 500 which leaves a remainder of 3 when it is divided by 7, which leaves a remainder of 4 when it is divided by 8, and which leaves a remainder of 5 when it is divided by 9.

### 3 Answers

- PuzzlingLv 73 months agoFavourite answer
Notice if we add 4 to the number, it will have *no* remainder when divided by 7, 8 or 9. Given those numbers have no factors in common (other than 1), the lowest common multiple of 7, 8 and 9 is 7 * 8 * 9 = 504

Subtracting 4 we get back to the original number 500 (which still matches the criterion of NOT being larger than 500).

Double-checking:

500/7 = 71 r 3

500/8 = 62 r 4

500/9 = 55 r 5

Answer:

500

- 3 months ago
7a + 3 = 8b + 4 = 9c + 5

7a = 9c + 2

7a = 8b + 1

8b = 9c + 1

7a - 2 = 9c

5 , 12 , 19 , 26 , 33 , 40 , 47 , 54 , 61 , 68 , 75 , ...

+5 , +3 , +1 , +8 , +6 , +4 , +2 , +0 , +7 , +5 , +3 , ....

9 * 6 = 7 * 8 - 2

Every 9 multiples of 6, we should get a number

7 * 17 - 2 = 117

7 * 26 - 2 = 180

7 * 35 - 2 = 243

7 * 44 - 2 = 306

7 * 53 - 2 = 369

7 * 62 - 2 = 432

7 * 71 - 2 = 495

Throw out the odd ones, since 8b + 4 will always be even

180 , 306 , 432

Throw out the one that isn't divisible by 4, since 8b + 4 is always divisible by 4

180 , 432

Throw out the one that is evenly divisible by 8

180