Number theory - Mixed bag - Part 1

• A positive whole number less than 100 is represented in base 2, base 3 and base 5 notations. It is found that exactly in two cases the last digit is 1 and in exactly two cases first digit is 1.What is the number in decimal representation?
a) 31
b) 61
c) 71
d) 81

In exactly two cases last digit is 1.This means that when the number is divided by two of this three numbers (2, 3, 5), the remainder should be 1.
Since the LCM of (2, 3, 5) is 30, numbers 31 and 61 give 1 remainder in all 3 cases.
Hence 31 and 61 can be eliminated.
71 in base 10 -> 1000111 in base 2 -> 2122 in base 3 -> 241 in base 5
81 in base 10 -> 1010001in base 2 -> 10000 in base 3 -> 311 in base 5
Hence only 81 have the first digit is 1 in two cases.

An Egyptian fraction has a numerator equal to 1 and its denominator is a positive integer. What is the maximum number of different Egyptian fractions such that their sum is equal to 1 and their denominators are equal to 10 or less?
a)3
b)5
c) 7
d) 9

We can ignore 1/7 and 1/9 because no sum of the other numbers is going to give 7th or 9th in the denominator. Also 1/5 and 1/10 are not enough to add up to anything since 1/10, 2/10, 3/10 are going to leave tenths left over no matter what else we add. Now whats left Is 1/2 , 1/3, 1/4, 1/6, 1/8 .The sum of these is 11/8 . So we need all of them except 3/8 ( 1/4 +1/8).
That means only way to do this is (1/2+1/3+1/6) .Hence maximum number of fractions needed is 3.

Sachin writes all numbers from 1 to 1000 on a paper in order. Find the 2883 rd digit written by him?

One digit numbers digits -> 9
Two digit numbers digits -> 180
Three digit numbers digits -> 2700
Total -> 2889
Last 3 digits will be 999…..then 998….hence 2883 rd digit will be 3.

If N = 72 then find the number of pairs (a,b) such that a and b are the factors of N and a * b = N?

N = 72
Factors are 1,2,3,4,6,8,9,12,18,24,36,72
If we form pairs (1,72) (2,36) ( 3,24) (4,18) (6,12) ( 8,9)
Total 6 pairs.

If N = 72 then find the number of pairs (a,b) for N such that a and b are the factors of N and are co prime to each other?

72 = 3^2 * 2^3
Total number of factors = 3 * 4 = 12
Factor 1 is co prime to all other factors. -> 11 pairs
Powers of 2 will be co prime with powers of 3
Powers of 2 -> 2,4,8 (3 values)
Powers of 3 -> 3,9 (2 values)
Hence 3 * 2 = 6 pairs possible.
So total 11+6 = 17 pairs possible.

The LCM of first 100 natural numbers is K .Find the LCM of first 105 natural numbers?

Given LCM of 100 natural numbers = K
To find the LCM of first 105 natural numbers, factorize the first 5 numbers after 100.
101 - prime
102 - 2 * 3 * 17
103 - prime
104 - 8 * 13
105- 3 * 5 * 7

during the calculation of LCM of first 100 natural numbers, the numbers 2, 3, 17, 13,8,5,7 are already considered. Hence only numbers which will affect the LCM will be 101 and 103 only.
Hence LCM = K * 101 * 103 = 10403K

Three thieves robbed a diamond shop independently but distributed their diamonds according to the rule that person or persons with maximum diamonds should give diamonds equal to HCF of their individual number of diamonds to the person or persons with the lowest number of diamonds. This process is repeated till all have equal diamonds.

Q1) If they have 39, 52, 26 diamonds ,how many times should the process be repeated?
a) 1
b) 3
c) 4
d) Not possible

Q2) If they have 51, 85, 102 diamonds how many times should the process be repeated?
a) 3
b) 4
c) 5
d) Not possible

a)HCF (39, 52, 26) = 13
first step, 13 will be deducted from second person and 13 will be added to third person.
Then it will become (39,39,39)

b)Total number of diamonds = 51 + 85 + 102 = 238
But its not a multiple of 3. So we can’t divide equally among three people.
Hence option 4) not possible is the answer.

Three containers having 4 ½ , 3 ¼ and 6 1/3 litres of three different cold drinks are to be served equally to all guests in a party such that each one gets maximum. How many maximum number of guests could be entertained?
a)50
b)150
c)169
d)196

HCF (9/2 , 13/4 , 19/3) = 1/12
Total number of guests = (9/2)/(1/12) + (13/4)/(1/12) + (19/3)/(1/12)
= 54 + 39 + 76 = 169

In a book store “OM INNOVATIVE BOOK STORE “is flashed using neon lights. The words are individually flashed at intervals of 2 2/3, 5 1/3,4 2/3,7 ½ seconds respectively and each word is put off after a second. The least time after which the full name of the book store can be read for a second is
a) 1734.6 seconds
b) 733.7 seconds
c) 3553 seconds
d) 2741.5 Seconds

Each word is put off after a second.
So OM will flash after 2 2/3 + 1 = 11/3 seconds.
Innovative will flash after 5 1/3 + 1 = 19/3 seconds
Book will flash after 17/3 seconds
Store will flash after 17/2 seconds
Time when they will flash again together next is LCM[ 11/3 , 19/3 , 17/3 , 17/2 ] =
= LCM [ 11,17,19 ] = 3553 SECONDS

Which one is bigger (9)^1/3 or (20)^1/4 ?

9^(1/3) = 9^(4/12) = (6561)^1/12
20^(1/4) = 20^(3/12) = (8000)^1/12
Hence 20^1/4 is greater.

When a number is divided by 147 the remainder is 50.What will be the remainder if the number is divided by 21?
a) 0
b) 5
c) 8
d) cannot determined

N = 147Q + 50
N mod 21 = 147Q MOD 21 + 50 mod 21 =0+50 mod 21 = 8

The factorial of a number can be decomposed in to a product of prime numbers.Let the maximum power of 2 in the decomposition of the factorial of a number (n) be k. Find n, if k = 9?
a) 10
b) 12
c) 13
d) no such n exists**

K = n/2 +n/4 +n/8……..
when n= 10 k = 5+2+1 = 8
when n= 12 k = 6+3+1 = 10
Hence no such number exists.

If P = 10 * 10! + 11 * 11! + 12 * 12! +……..99 * 99!
Then p will end in how many zeros?
a) 1
b) 2
c) 24
d) 25

10!(11-1) +11! *(12-1) + 12!(13-1)…..
=11!-10!+12!-11!+13!-12!...........
= 100! – 10!
Hence it will end in 10/5 = 2 zeros.

If X = 13^3 +14^3 +15^3+16^3 then what is the remainder when x is divided by 58?
a) 0
b) 1
c) 57
d) 29

A^n + B^n is divisible by A+B when n is odd.
So x is divisible by 29
X= 29(13^2+16^2-13 * 16) + 29(14^2+15^2-14 * 15)
= 29(13^2+14^2+15^2+16^2-13 * 16 - 14 * 15) = 29 * an even number
Hence x is a multiple of 58..so answer is 0.

Find the greatest number consisting of 6 digits which on being divided by 6,7,8,9,10 leaves 4,5,6,7,8 as remainders respectively.
a) 997920
b) 997918
c) 999999
d) 997922

LCM [6,7,8,9,10] = 2^3 * 3^2 * 7 * 5 = 2520
Greatest 6 digit number which is a multiple of 2520 = 997920
Required number = 997920 – 2 = 997918.
PS : we are subtracting 2 coz 6-4 = 7-5 = 8-6 = 9-7 = 10-8 = 2

Find the least number which being divided by 5,6,8,9,12 leaves in each case a remainder 1 but when divided by 13 leaves no remainder?
a) 2987
b) 3601
c) 3600
d) 2986

LCM [5,6,8,9,12] = 360
360A +1 = 13B
From option the number of the form 360+1 is 3601 only.

94^3 – 23^3 - 71^3 is at-least divisible by?
a) 71 and 23
b) 23 and 74
c) 71 and 94
d) 23, 71 and 94

94^3-23^3 is divisible by 94-23 = 71
94^3-71^3 is divisible by 94-71 =23
23^3+71^3 is divisible by 23+71 = 94
Hence it is divisible by 23, 71 and 94.

The last digit of a is same as the last digit of its square. How many such numbers exists from 0 to 90?
a) 34
b) 37
c) 9
d) 4

Unit digit of a number and its square will be equal when unit digit is any of 0,1,5,6.
From 0 to 90 : unit digit 0 ->10 numbers
From 0 to 90 : unit digit 1,5,6 -> 9 each
Hence total 37 numbers.

N = 2^15 * 3^12 How many factors of N^2 are less than N but not divide N completely?

Factors of N^2 = 31 * 25 = 775
Factors of N^2 less than N = 775 * 1/2 = 387
Factors of N = 16 * 13 = 208
hence answer is 387-207 = 180.

If a number is divided by M gives a remainder 17, when same number is divided by (8 * M) remainder comes 47. Then what would be the remainder when same number is divided by (4 * M)?

MQ+17 = 8Mq+47
M (Q-8q) = 30
M > 17
so M will be 30.
Q-8q = 1
let Q = 9 and q = 1
Then number is 287.
287 mod 120 = 47.

Let N = 2^3 * 3^4 * 5^9. Then find the product of all factors of N which are not divisible by 12?

Product of factors of N = N^ (f/2) where f is number of factors.
Number of factors f = 4 * 5 * 10 = 200
Product of number of factors = 2^300 * 3^400 * 5^ 900
Product of factors which are divisible by 12 -> 2 * 3^3 * 5^9
f = 2 * 4 * 10 = 80
product = 2^40 * 3^120 * 5^360 * 12 ^80
Required product = (2^300 * 3^400 * 5^900)/2^40 * 3^120 * 5^360 * 12^80.
= 2^100 * 3^200 * 5^540.

There are 47 identical coins. All the coins have same weight except one coin which has a different weight. Whats the minimum number of weighing required to be certain of identifying the coin with different weight?

If N is the number of balls and question doesn’t specify whether the ball weighs more or Less , ten total number of weighing = m+1
Where N < = 3^m+3^(m-1) +3^(m-2) +……..+3^0
here N = 47
47 < = 1+3+9+27+81
hence m = 4
so number of weighing required 4+1 = 5.

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