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
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4 Answers

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 42 833.3333 49 714.2857 56 625 63 555.5556
35001 42 833.3571 49 714.3061 56 625.0179 63 555.5714
35002 42 833.381 49 714.3265 56 625.0357 63 555.5873
35003 42 833.4048 49 714.3469 56 625.0536 63 555.6032
35004 42 833.4286 49 714.3673 56 625.0714 63 555.619
35005 42 833.4524 49 714.3878 56 625.0893 63 555.6349
35006 42 833.4762 49 714.4082 56 625.1071 63 555.6508
35007 42 833.5 49 714.4286 56 625.125 63 555.6667
35008 42 833.5238 49 714.449 56 625.1429 63 555.6825
35009 42 833.5476 49 714.4694 56 625.1607 63 555.6984
by Top Rated User (103k points)
3528 lcm the 3528-3500 =28
by

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
ago by

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