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Let x be the least number of 4 digits that when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder of 1 in each case. If x lies between 2000 and 2500, then what is the sum of the digits of x?

A15

B4

C9

D10

Answer:

B. 4

Read Explanation:

LCM of 2, 3, 4, 5, 6 and 7 is 420. Number should be between 2000 and 2500 so, the number is = 420 × 5 = 2100 Remainder will be 1 in each case so the number is = 2100 + 1 = 2101 Sum of digit of the number 2101 is = 2 + 1 + 0 + 1 = 4


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A=23×35×52,B=22×3×72A=2^3\times3^5\times5^2,B=2^2\times3\times7^2

$$find the HCF of A & B