which last number should be added to 3500 to make it exactly divisible by 42,49,56 and 63

Question

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

Answer 1

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

Answer 2

3528 lcm the 3528-3500 =28

Answer 3

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.

Answer 4

3528