161.. Required number = (555 + 445) x 2 x 110 + 30 = 220000+30 =220030.
162. Largest number of 4 digits = 9999. On dividing 9999 by 7, we get remainder = 3.
Largest number of 4 digits divisible by 7 is (9999 by 7, we get remainder = 3.
Let (9999 – X + 10) be divisible by 3. By hit and trial , we find that X = 7.
Required number = (9996 – 7)= 9989.
163. Number = (144 x Q) + 19 x 6 Q + 19 + 2 = 19 x (6 Q + 1) +1.
Required remainder =2.
164. Number = (296 x Q) + 75 = (37 x 8 Q) + 37 x 2) + 1= 37 x (8 Q +3) +1
Required remainder = 1.
165. Number = (199 x Q) + 29 = (29 x 31 x Q) + ( 20 x )31 Q + 2) + 5.
Required remainder = 5.
166. Number = (5 x 46) = 230. Also, 10 x Q =230 Q =23. And, R =46.
Dividend = (230 x 23 +46) =5336.
167. Let the smaller number be X then, larger number = (1365 + X).
126. 1365 + X = 6 X + 15 5 X =1350 X=270.
Hence, the required number is 270.
168. Dividend = (12 x 35) =420. Now, divided = 420 and divisor = 21.
Correct quotient = 420/2120.
169. Let n = 4 q + 3 2 n = 8 q +6=(8 q +4) + 2 2 n = 4 (2 q + 1) + 2.
So, when 2 n is divided by 4, remainder =2.
170. Let x = 6 q + 3. Then, X² (6 q +3 )² = 36 q² + 36 q + 9 = 6(6 q² +6 q + 1) + 3.
So, when X² is divided by 6, remainder =3.
171. 4 x
5 y – 1
1 -4
Y = (5 x 1 + 4) = 9
X = (4 Y + 1) = (4 x 9 + 1) = 37
Now 37 when divided successively by 5 and 4, we get:
5 37
4 7 – 2
1 – 3
Respective remainders are 2, 3.
172. 
4 X
4 X
5 y - 2
6 1 - 4
173. 
5 X
5 X
9 
y - 4
y - 4
13 Z -8
1- 12
Now, 1169 when divided by 585 gives remainder = 584.
174. Let n = 3 q + 1 and let 2 p + 1 + 1 + 6 p + 4. Then, n = 3(2 p + 1 ) + 1 = 6 p + 4
The number when divided by 6, we get remainder = 4.
175. 4⁶¹ + 4⁶² + 4⁶³ + 64 ( 1 + 4 + 4 ² + 4³) 4⁶¹ x 85 = 4⁶⁰x 430, which is clearly divisible by 10.
176. Putting x = 2, we get 2² (2² - 1) = 12. So, X² (X² - 1) is always divisible by 12.
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