Question
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. The sum of the digits in N is:
Answer: Option A
:
A
Since the remainder is same in each case, hence we will use the following formula
H.C.F (x, y, z) = H.C.F of (x -y), (y- z), ( z-x)
N = H.C.F. of (6905 - 4665), (4665 - 1305), and (6905 - 1305)
= H.C.F. of 2240, 3360and 5600
HCF of 5600 and 3360:
5600=3360×1+2240
3360=2240×1+1120
2240=1120×2+0
∴ HCF of5600 and3360= 1120
Now, HCF of 2240 and 1120 = 1120
So, the HCF of (3360, 2240 and 5600) = N = 1120
Sum of digits in N = (1 + 1 + 2 + 0) = 4
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:
A
Since the remainder is same in each case, hence we will use the following formula
H.C.F (x, y, z) = H.C.F of (x -y), (y- z), ( z-x)
N = H.C.F. of (6905 - 4665), (4665 - 1305), and (6905 - 1305)
= H.C.F. of 2240, 3360and 5600
HCF of 5600 and 3360:
5600=3360×1+2240
3360=2240×1+1120
2240=1120×2+0
∴ HCF of5600 and3360= 1120
Now, HCF of 2240 and 1120 = 1120
So, the HCF of (3360, 2240 and 5600) = N = 1120
Sum of digits in N = (1 + 1 + 2 + 0) = 4
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