Simple problems:
1.What least number must be added to 3000 to obtain a number
exactly divisible by 19?
Solution:
On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.
2.Find the unit's digit n the product 2467 153 * 34172?
Solution:
Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.
3.Find the total number of prime factors in 411 *7 5 *112 ?
Solution:
411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29
4.The least umber of five digits which is exactly
divisible by 12,15 and 18 is?
a.10010 b.10015 c.10020 d.10080
Solution:
Least number of five digits is 10000
L.C.Mof 12,15,18 s 180.
On dividing 10000 by 180,the remainder is 100.
Therefore required number=10000+(180-100)
=10080.
Ans (d).
5.The least number which is perfect square and is divisible
by each of the numbers 16,20 and 24 is?
a.1600 b.3600 c.6400 d.14400
Solution:
The least number divisible by 16,20,24 = L.C.M of 16,20,24=240
=2*2*2*2*3*5
To make it a perfect square it must be multiplied by 3*5.
Therefore required number =240*3*5=3600.
Ans (b).
6.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.
The positive integer is?
a.1 b.3 c.5 d.7
Solution:
1000+N>1000N
clearly N=1.
7.How many numbers between 11 and 90 are divisible by 7?
Solution:
The required numbers are 14,21,28,...........,84.
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.
8.Find the sum of all odd numbers up to 100?
Solution:
The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.
9.How many terms are there in 2,4,6,8..........,1024?
Solution:
Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.
10. 2+22+23+24+25..........+28=?
Solution:
Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.
11.Find the number of zeros in 27!?
Solution:
Short cut method :
number of zeros in 27!=27/5 + 27/25
=5+1=6zeros.
Medium Problems:
12.The difference between two numbers 1365.When the larger
number is divided by the smaller one the quotient is 6 and
the remainder is 15.The smaller number is?
a.240 b.270 c.295 d.360
Solution:
Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.
13.Find the remainder when 231 is divided by 5?
Solution:
210 =1024.
unit digit of 210 * 210 * 210 is 4
as 4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.
14.The largest four digit number which when divided by 4,7
or 13 leaves a remainder of 3 in each case is?
a.8739 b.9831 c.9834 d.9893. Solution:
solution:
Greatest number of four digits is 9999
L.C.M of 4,7, and 13=364.
On dividing 9999 by 364 remainder obtained is 171.
Therefore greatest number of four digits divisible by 4,7,13
=9999-171=9828.
Hence required number=9828+3=9831.
Ans (b).
15.What least value must be assigned to * so that th number
197*5462 is divisible by 9?
Solution:
Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.
16.Find the smallest number of 6 digits which is exactly
divisible by 111?
Solution:
Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.
