Question 121

Let x be the least 4-digit number which when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder of | in each case. If x lies between 2800 and 3000, then what is the sum of the digits of x?

Solution

Given numbers (2,3,4,5,6,7)

taken Lcm = 420 from given above numbers

Now, multiply of 420 are $$ 420 \times 1 = 420 $$

$$\Rightarrow 420 \times 2 = 840 $$

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$$420 \times 7 = 2940 +1 = 2941 $$ (which is lies between 2800 and 2900)

then sum of digits = 2+9+4+1 = 16 Ans

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