# 2IIM Quant Notes - Number Theory - Part 2

• How many numbers with distinct digits are possible product of whose digits is 28?
6
4
8
12

Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74.
Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.
We cannot have three digits as (2, 2, 7) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.

A page is torn from a novel. The sum of the remaining digits is 10000. What is the sum of the two page-numbers on the torn page of this novel?
77 and 78
33 and 34
8 and 9
None of these

n (n + 1)/2 should be nearby 10,000, so n (n + 1) is somewhere near 20,000. So n should be around √20000 about 141.
Try 142
142 * 143 = 10153
141 * 142 = 10011
So, the missing pages are either 76 and 77 or 5 and 6.

4-digit number of the form aabb is a perfect square. What is the value of a - b?
3
2
4
1

A number of the form aabb has to be a multiple of 11. So, it is the square of either 11 or 22 or 33 or...so on up to 99.
88^2 = 7744.
This is the only solution possible. Most of these trial and error questions need to be narrowed down a little bit before we can look for the solution. That narrowing down is critical. In this case, we should look for multiples of 11.

What are the last two digits of the number 7^45?
07
23
49
43

The last two digits of 7^1 are 07.
The last two digits of 7^2 are 49.
The last two digits of 7^3 are 43.
The last two digits of 7^4 are 01.
The last two digits of powers of 7 go in a cycle - 07, 49, 43, 01.
So, the last two digits of 7^45 are 07.

n^2 + 5n + 6 is a multiple of 6. n is natural number less than 100. How many values can n take?
33
65
66
67

n^2 + 5n + 6
(n + 2) (n + 3)
(n + 2) and (n + 3) are two consecutive numbers.
One of (n + 2) or (n + 3) has to be even.

We need n such that (n + 2) (n + 3) is a multiple of 3.
(n + 2) or (n + 3) should be a multiple of 3.
n leaves a remainder of 0 or 1 when divided by 3.
n could be 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18 ....
So, from among the first 99 natural numbers, N cannot take 2, 5, 8, 11, 14…….98.
There are 33 numbers in this list –
we are effectively listing all numbers {3 × 0 + 2, 3 × 1 + 2, 3 × 3 + 2, ….3 × 32 + 2}.
So, N can take 99 – 33 = 66 values.

[x] is the greatest integer less than or equal to x. Find the number of positive integers n such that .
31
25
35
40

This is a classic case of brute-force counting.
When n = 1, 2, 3....10, both values will be equal to 0 - 10 possibilities
When n = 13, 14, ....21, both values will be equal to 1 - 9 possibilities
When n = 26, 27, ....32, both values will be equal to 2 - 7 possibilities
When n = 39, 27, ....43, both values will be equal to 3 - 5 possibilities
When n = 52, 53, 54, both values will be equal to 4 - 3 possibilities
when n = 65, both will be equal to 5 - 1 possibility
So, totally there are 1 + 3 + 5 + 7 + 9 + 10 = 35 possibilities.

Positive numbers 1 to 55, inclusive are placed in 5 groups of 11 numbers each. What is the maximum possible average of the medians of the 5 groups?
34
28
35
38

We need to maximise each median in order to have the overall maximum median possible.

The highest possible median is 50 as they should be 5 numbers higher than the median in the group of 11. So, if we have a set that has a, b, c, d, e, 50, 51, 52, 53, 54, 55, the median will be 50. In this set, it is best not to waste any high values on a, b, c, d or e as these do not affect the median. So, a set that reads as 1, 2, 3, 4, 5, 50, 51, 52, 53, 54, 55 will also have median 50.

The next set can be 6, 7, 8, 9, 10, 44, 45, 46, 47, 48, 49. The median will be 44.

The medians of the 5 groups can be 50, 44, 38, 32, 26. The highest possible average of the medians will be 38.

N is an 80-digit positive integer (in the decimal scale). All digits except the 44th digit (from the left) are 2. If N is divisible by 13, find the 26th digit?
5
6
1
2

To begin with, the question should read "find the 44th digit".
Any number of the form abcabc is a multiple of 1001. 1001 is 7 * 11 * 13. So, any number of the form abcabc is a multiple of 13.
So, a number comprising 42 2's would be a multiple of 13, so would a number comprising 36 2's. So, in effect, we are left with a two digit number 2a, where a is the 44th digit. 26 is a multiple of 13, so the 44th digit should be 6.

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.