Numbers – II

x-6 = 11 or x-6 =-11 -> x = 17 or -5	2y-12 = 8 or 2y-12 = -8 -> y = 10 or 2
Minimum = 10/-5 = -2 Note: Maximum value will be 10/17.

Sum of the first n natural numbers = n(n+1)/2

Here, n = 100 -> Sum = (100*101)/2 = 50*101 = 2*25*101

Now, 101 is a prime number and out of 25 and 2, 25 is divisible by only 5 and 25. Hence, the sum is divisible by only 2 (out of the given options)

Handle the question through understanding a pattern evolved by taking experimental values.

(2³-1³)/(2²+1²) = 7/5 > 1

(3⁴-2⁴)/(3³+2³) = (81-16)/(27+8) = (65/35) > 1

In general, [n^m-(n-1)^m]/[n^m-1+ (n-1)^m-1] > 1

241! = 241*240*239*…..1

When we divide 241! by a power of 5, we have 241 numbers in the numerator. The denominator will have all 5s.

The 241 numbers in the numerator have 241/5 = 48.2 or 48 multiples of 5 i.e. 5, 10, 15, 20, 25 ….. 240.

Corresponding to each of the above multiples, we have a 5 in the denominator which will divide the numerator completely.

Therefore, 5⁴⁸ can definitely divide 241!

Further, every multiple of 5² i.e. 25, after cancelling out 5 as above, will have one more 5 left. Therefore, for every such multiple of 25, we have an additional 5 in the denominator. There are 241/25 = 9.64 or 9 such multiples of 25. So, we can take 9 more 5s in the denominator.

Similarly, for every multiple of 5³ i.e. 125, we can take an additional 5 in the denominator. There are 241/125 = 1.928 or 1 such multiple of 125.

Total number of 5s= 48+9+1 = 58

So, 5⁵⁸ divides 241!

Alternatively

Quicker Method

5	241
5	48
5	9
	1

Number of 5s =48+9+1 = 58

This method is applicable only if the number whose largest power is to be found is a prime number. In case the number is not prime, then express the number as a product of two numbers which are co-prime; apply the concept to each of these two numbers

Applying an approach similar to Q4 (refer to the note: 2 is a prime number, hence this approach may be applied), we get the largest power of 2 by the following calculation:

2	100
2	50
2	25
2	12
2	6
2	3
	1

 

Number of 2s= 50+25+12+6+3+1 = 97

Here, 10 is not a prime number; so, we express 10 as a product of two numbers which are co-prime (refer to the note in Q4) Now, 10 = 2*5

2	250
2	125
2	62
2	31
2	15
2	7
2	3
	1

Largest power of 2 =125 + 62 + 31 + 15 + 7 + 3 + 1 = 244

5	250
5	50
5	10
	2

Number of zeros is determined by the number of 5s, because a 5 multiplied by an even digit will result in a 0. Number of 5s is determined by applying the approach used in Q4.

5	1221
5	244
5	48
5	9
	1

à Number of 5s = 244+48+9+1 = 302

Sum of the squares of the first 30 even numbers =

2²+4²+6²+ —– 60²  = 2²[1²+2²+3²+ —– 30²]

This is 4 times the sum of the squares of the first 30 natural numbers. Now, sum of the squares of the first n natural numbers = [n(n+1)(2n+1)]/6

Hence, required sum = 4*[(30*31*61)/6] = 37820

Sum of the cubes of the first 20 odd numbers =

1³+3³+5³+ —– 39³ = (1³+2³+ —– 40³) – (2³+4³+ 6³—– 40³)

= (1³+2³+ —– 40³) – 2³[1³+2³+ —– 20³]

Now, sum of the cubes of first n natural numbers is [n(n+1)/2]²

= [(40*41)/2]² – 8[(20*21)/2]²

= 672400-352800 = 319600

x and y can both be either positive or negative, leading to the following cases:   x y x-y |x| |y| |x|-|y| |x-y| 6 4 2 6 4 2 2 6 -4 10 6 4 2 10 -6 4 -10 6 4 2 10 -6 -4 -2 6 4 2 2 4 6 -2 4 6 -2 2 4 -6 10 4 6 -2 10 -4 6 -10 4 6 -2 10 -4 -6 2 4 6 -2 2   In all cases |x|-|y| ≤ |x-y| Hence, option C