# Question Bank - Number Theory

• (abc) base 9= (cba) base 11
a * 81 + b * 9 + c = c * 121 + b * 11 + a
a,b,c must be less than 9 and a,c can not be 0.
a * 81 + b * 9 + c = c * 121 + b * 11 + a
120c - 80a + 2b=0
60c - 40a + b = 0
c= 2, a=3, b=0 satisfies
You can also check
(302)9= (203)11
But c=4 and a=6 will also satisfy.
Also c= 6 and a=9, but a,b,c can not be 9, so ignore.
So answer is CBD.

• Q26) How many 2 digits number in base 9 gets reversed when expressed in base 11

• (ab)9= (ba)11
9a+b= 11b+a
8a=10b
a/b= 5/4
a and b are single digit so only choice is 5,4.
So 1 number.

• Q27) Number of zeroes in 6! In base 6.

• Number of trailing zeroes in base n: So in decimal system we used to check for highest power of 10, so in any base system we will check for the highest power of n.
Example : Number of zeroes in 6! In base 2
6/2 + 6/4 = 3 + 1 = 4

Number of zeroes in 6! In base 6.
6 = 3 * 2
So check by 3 and 2. But we know 3 will be less so check directly by 3.
6/3=2
So 2 zeros

• Q28) What is the remainder when 47^37^27 is divided by 11?

• Q29) How many numbers between 0 and 500 are divisible by 3 or 5 or 7?

• The question can be solved using the concept of set theory,
Numbers that are divisible by 3, 5 and 7 can be represented as n(3), n(5) and n(7) respectively.
Between 0 to 500,
n(3) = 500/6= 166
n(5) = 500/5 = 100
n(7)= 500/7 = 71
Total numbers = n(3) + n(5) +n(7)= 166+100+71 = 337

Here, we have calculated numbers that are divisible by 3,5 and 7 individually.
There are some numbers that have been repetitively counted, in order to eliminate them we have to find out the common numbers.

Numbers that are divisible by 3 and 5 = n(3&5) = 500/ (LCM of 3 & 5) = 500/15 = 33
Numbers that are divisible by 5 and 7 = n(5&7) = 500/ (LCM of 5 & 7) = 500/35 = 14
Numbers that are divisible by 3 and 7 = n(3&7) = 500/ (LCM of 3 & 7) = 500/21 = 23
Numbers that are divisible by 3,5 and 7 = n(3&5&7) = 500/(LCM of 3&5&7) = 500/ 105 = 4

Formula of set theory can be used,
n(A or B or C) = n(A)+n(B)+n(C)-n(A&B)-n(B&C)-n(A&C)+n(A&B&C)
Substituting the values in the given formula, we obtain
The actual numbers that are divisible by 3 or 5 or 7 = 337–33–14–23+4 = 271

[solved by vivek subramanian]

• Q30) M is a two digit number such that the product of the factorials of its individual digits is greater than the sum of the factorials of its individual digits. How many values of M exist?
a) 56
b) 64
c) 63
d) None of these

• Approach: Numbers having 1/0 in them will not satisfy
a! * b! > a! + b! as 1 * something is always less than 1 + something.
So eliminate numbers having 1/0 in them
11-91, 10-90, 12-19= 26 numbers
Also, an exception is 22 where 2! + 2! = 2! * 2!
Hence, total 27 numbers will not satisfy.

So, 90-27=63 numbers will satisfy

[solved by vinit sanghvi]

58

30

41

54

34

47

53

47