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H.C.F and L.C.M Problems-Data Intensity 1 :: Discussion
| 4. | Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is? | |
| A. | 4 | |
| B. | 5 | |
| C. | 6 | |
| D. | 8 | |
| Answer: Option A | |
| Explanation: | |
|
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4 |
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