111. 2² +4² X ……+ 20²=(1x2)² + (2 x 2)² + (2 x 3)² + …. + (2 x 10)²
=2² x 1² + 2² x 2² + 2² +3² +…. + 2² x 10²
= 2² x [1² + 2² x+ 3³+….+ 10²]
=4 x 10 x 11 x 21 =4x 385 = 1540.
6
112. 11² + 12² + 13² + ….+ 20²
= ( 1² + 2² + 3² + ….. + 20²) –(1² + 2²+ 3² + …. + 10²)
=[ 20 (20 + 1 ) ( 40 + 1) - (10 + 1) (20 +1)] = 2485.
6 6
113. 1 + X + 5 + 4 + 8 = (18 + x). Clearly, when X=.0, then sum of digits is divisible by 3.
114. Let the required number be 375 y 25X.
Then, for divisibility by 5, we must have X =5.
Case 1. When X = 0.
Then, sum of digits = (22 + y ). For divisibility by 3, (222 + y) must be divisible by 3.
Y = 2 or 5 or 8.
Numbers (0, 2) or ( 0, 5) or (0, 8).
Case II. When x +=5.
Then, sum of digits = (27 + y). For divisibility by 3, must have by =0 or 3 or 6 or 9.
Number are (5,0) or (5, 3 ) or (5, 6) or ( 5, 9).
So, correct answer is (b).
115. Let the number be 5X2. Clearly, it is divisible by 2.
Now, 5 + X + 2 = (7 + X) must be divisible by 3. So, x = 2.
116. The given number is divisible by 8, if the number 6X2 is divisible by 8.
Clearly, the lest value of X is 3.
117. (4 + 5 + 1 + X + 6 + 0 + 3) =19 + X. clearly, X = 8
118. Taking the sum of the digits, we have:
119. s₁ = 9, s₂ = 12, s₃ =18, s₄ =9, s₅=21, s₆=12, s₇ =18, s₈=21, s₉ =15, s₁₀ =24.
Clearly, s₁, s₂, s₃, s₄, s₅, s₆, s₇, s₈, s₉, s₁₀ are all divisible by 3 but not boy 9.
So, the number of required numbers = 6.
120. (a) ( 1+ 6 + 3 ) – ( 2 + 5 + 4) = 1 (No) (b)( 2 + 6 + 4) – (4 + 5 + 2) =1(no)
(c) (4 + 6 + 1 ) – ( 2 + 5 + 3) = 1 ( no) (d) ( 4+ 6 + 1 ) – ( 2 + 5 + 4) =0 (yes).
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