how can i solve this question which last number should be added to 3500 to make it exactly divisible by 42,49,56 and 63

You might want to explain what you mean by "exactly divisible."  If you mean "divisible with no remainder, the answer having no digits after the decimal point," then:

From the chart below, there is no solution for 42, 49, and 63.

If you add 0 to the end of 3500, you will find that 35000 is exactly divisible by 56.

 35000 42 833.333 49 714.286 56 625 63 555.556 35001 42 833.357 49 714.306 56 625.018 63 555.571 35002 42 833.381 49 714.327 56 625.036 63 555.587 35003 42 833.405 49 714.347 56 625.054 63 555.603 35004 42 833.429 49 714.367 56 625.071 63 555.619 35005 42 833.452 49 714.388 56 625.089 63 555.635 35006 42 833.476 49 714.408 56 625.107 63 555.651 35007 42 833.5 49 714.429 56 625.125 63 555.667 35008 42 833.524 49 714.449 56 625.143 63 555.683 35009 42 833.548 49 714.469 56 625.161 63 555.698
by Top Rated User (103k points)
3528 lcm the 3528-3500 =28
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42, 49, 56 and 63 are all divisible by 7 and 3500 is also divisible by 7. So we need to find the smallest number we can add to 3500/7=500 to make a number that’s divisible by the LCM of 42/7, 49/7, 56/7 and 63/7, that is, the LCM of 6, 7, 8 and 9. These numbers can be written 2×3, 7, 2×4 and 3×3, so the LCM=2×3×3×4×7=504. The number to add to 500 is therefore 4.

So 504 is divisible by 6, 7, 8 and 9. Therefore 7×504=3528 is divisible by 42, 49, 56 and 63. So the least number to be added to 3500 is 28, giving us 3528.

by Top Rated User (781k points)
3528
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