Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.
Let d and s represent the number of daughters and sons respectively.
Then, we have :
d - 1 = s and 2 (s - 1) = d.
Solving these two equations, we get: d = 4, s = 3.
Total number of digits
= (No. of digits in 1- digit page nos. + No. of digits in 2-digit page nos. + No. of digits in 3- digit page nos.)
= (1 x 9 + 2 x 90 + 3 x 267) = (9 + 180 + 801) = 990.
Total runs scored = (36 x 5) = 180.
Let the runs scored by E be x.
Then, runs scored by D = x + 5; runs scored by A = x + 8;
runs scored by B = x + x + 5 = 2x + 5;
runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.
So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.
Therefore 3x + 120 =180 ⇔ 3X = 60 ⇔ x = 20.
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