# Quant Boosters - Swetabh Kumar - Set 2

• Q25) If N be the number of consecutive zeros at the end of the decimal representation of the expression 1! × 2! × 3! × 4! x ... x 99! × 100! Find the remainder when N is divided by 1000?

• 1^100 * 2^99 * 3^98 ... 100^1
power of 5: (1+6+11....96)+(76+51+26+1) = 970+154=1124
so 1124 mod 1000=124

• Q26) A and B started running from the same point and in opposite directions around a circular track of radius 24.5 m. A’s speed was twice that of B’s speed. They met each other after 14 seconds. What was A’s speed?

• A=2x, B=x
length= 2pi * r = 49pi = 154 m.
so given, 154/3x = 14 x=11/3 so 2x=22/3

• Q27) Find the approximate sum of the series 11/4, 31/8, 79/16 up to 10 terms?

• (2+1-1/4) + (3+1-1/8) + (4+1-1/16)...10 terms
(2+3+4....11) + (1+1+1...10 times) - (1/4+1/8+1/16...10 terms of GP)
= 65 + 10 - 1/2{1-(1/2)^10} = 75- 1/2 {1023/1024} = 74.5 approx

• 1 comes 3 times (from rt 1 to rt 3)
2 comes 5 times (from rt 4 to rt 8 )
3 comes 7 times (from rt 9 to rt 15) and so on.
so ever number N comes 2n+1 times. so N(2N+1)
till 43. After that the remaining 69 numbers with 44.

1 * 3 + 2 * 5 + 3 * 7..... 43 * 87 + 44 * 69
so 2n^2 + n sigma till 43 + 44*69
2 * 43 * 44 * 87/6 + 43 * 22 + 44 * 69 = 58850

• Q29) Given N = 35 x 36 x 37 .......... x 67, what is the remainder left when N is divided by 289?

• Q30) Find the number of integer solutions for |x| + 2|y| + |z| = 4

• 35 * 36 * ...67 , cancel one 17 with 51.
so remaining: (35 * 36....50) * 3 * (52 * 53....67) mod 17
(16!) * 16! * 3 mod 17 = 1 * 1 * 3 = 3 mod 17 (using Wilson)
since we cancelled 17, so 17 * 3= 51.

• |x|+2|y|+|z|=4
for |y|=0, |x|+|z|=4 so 4n=4 * 4=16 cases
for |y|=1, y=+-1 and |x|+|z|=2 so 2 * 4 * 2=16 cases
for |y|=2 y=+-2 |x|+|z|=0 so 2 * 1 = 2 cases
so 16+16+2 = 34 integer solutions.

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