# Quant Capsules by Shashank Prabhu - Set 17

• Find the sum of all numbers between 200 and 1500 that cannot be written as a sum of two or more consecutive numbers?

• Any odd number can be expressed as a sum of two consecutive integers
• Any even number that is in the form of 2^a * x^b * y^c... (in short, an even number not in the form of 2^n) can be expressed as a sum of three or more consecutive integers
• The only type that remains is numbers in the form of 2^n. These cannot be expressed as a sum of at least 2 consecutive integers

So power of 2 lying between 200-1500 => 256+512+1024 =>1792

There are 10 students out of which three are boys and seven are girls. In how many different ways can the students be paired such that no pair consists of two boys?

3 boys can be treated as 3 different units. Now, the girl corresponding to the first boy can be chosen in 7 ways, the girl corresponding to the second boy can be chosen in 6 ways and the girl corresponding to the third boy can be chosen in 5 ways. The remaining 4 girls can be teamed in 4c2/2!=3 ways.
Total 7 * 6 * 5 * 3 = 630 ways.

A natural number is written on each face of a cube so that the sum of the numbers on all the faces is S. A small triangular portion is sliced off from each corner of the cube. The product of the numbers on the faces that meet at a particular corner is written on the portion sliced off at that corner. The sum of the numbers written on all the sliced off portions is 2004. How many different values can S take?

abc, acd, ade, abe, bcf, cdf, def, bef will be the trios.
ac(b+d)+ae(b+d)+cf(b+d)+ef(b+d)=2004
{a(c+e)(b+d)+f(b+d)(c+e)}=2004
(a+f)(b+d)(c+e)=2004
2014=2^2 * 3 * 167
So split as (2,2,501) (2,6,167) (4,3,167) (2,3,334)

The sum of the squares of the first 15 positive integers (1^2 + 2^2 + 3^2 + . . . + 15^2) is equal to 1240. What is the sum of the squares of the second 15 positive integers (16^2 + 17^2 + 18^2 + . . . + 30^2) ?
(A) 2480
(B) 3480
(C) 6785
(D) 8215
(E) 9255

[30 * 31 * 61 - 15 * 16 * 31]/6
5 * 31 * 61 - 5 * 8 * 31
31(305 - 40)
31 * 265 = 8215

B takes 12 more days than A to finish a piece of work. B and A starts this work and A leaves the work 12 days before the work is finished. B completes 60% of overall work.How long would have B taken to finish the work if he works alone?

A works for x days, B works for x+12 days. A does 40% work in x days. So, A does 100% work in 2.5x days. B does 100% work in 2.5x+12 days. So, B does 60% work in 1.5x+7.2 days.
x+12=1.5x+7.2
0.5x=4.8
x=9.6
B does 100% work in 36 days

If both the roots of quadratic equation ax2 + bx + c =0 lie in the interval (0,3) then a lies in
(1) (1,3)
(2) (-1,3)
(3) (-1,3)
(4) (-121/91, -8)
(5) None of these

For any equation that has roots say (3,3) we can say that it will be in the form x^2-6x+9=0 or 2x^2-12x+18=0 or 3x^2-18x+27=0 and so on. As there is no cap on the value of a, none of the options can be said to be correct, I suppose.

In how many ways can we get a sum greater than 17 by throwing six distinct dice ?

6^6 total cases.
a+b+c+d+e+f 11c6 solutions. Similarly, b, c, d, e, f can be assumed to be equal to at least 6 and we will get the same number of cases. So, 6 * 11c6 negations in total.
Remaining cases = 6^6 - (17c6-6 * 11c6) = 37052.

Find the sum of all the integers N > 1 such that the each prime factor of N is either 2, 3 or 7 and N is not divisible by any perfect cube greater than 1.

The largest such number is 2^2 * 3^2 * 7^2
Now, the powers can vary from 0 to 2 in each case. This translates into simply finding the sum of all the factors of 2^2 * 3^2 * 7^2. Direct formula gives us 7 * 13 * 57=5187. But this includes 1 as well. So, sum of all such numbers > 1 will be 5186

What is the digit at the ten’s place of the number N = (6^11)^7

A lot of ways of doing this. Probably the most non-invasive one is to check for a pattern.
6^1 = 06
6^2 = 36
6^3 = 16
6^4 = 96
6^5 = 76
6^6 = 56
6^7 = 36 and it will continue.
So, for every term in the form of
5k -> 7
5k+1 -> 5
5k+2 -> 3
5k+3 -> 1
5k+4 -> 9
As 11^7 will be in the form of 5k+1, 5 it is.

What is the highest possible value of n for which 3^1024 – 1 is divisible by 2^n
(1) 13
(2) 10
(3) 11
(4) 12

(3^512+1)(3^512-1)
(3^512+1)(3^256+1)(3^256-1)
(3^512+1)(3^256+1)(3^128+1)(3^128-1)
(3^512+1)(3^256+1)(3^128+1)(3^64+1)(3^64-1)
(3^512+1)(3^256+1)(3^128+1)(3^64+1)(3^32+1)(3^32-1)
(3^512+1)(3^256+1)(3^128+1)(3^64+1)(3^32+1)(3^16+1)(3^16-1)
.
.
.
(3^512+1)(3^256+1)(3^128+1)(3^64+1)(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3^1+1)(3^2-1)
3^odd powers will give 3^x+1 in the form of 4k and 3^even powers will give 3^x+1 in the form of 2k.
So, total of 13 twos.

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