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<body bgcolor="aqua"><center><h2>Problems on Numbers</h2></center>
<pre>
<font size="2">
<strong>Simple problems:</strong>
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.
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<b>Medium Problems:</b>
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.
17.A number when divided by 342 gives a remainder 47.When
the same number is divided by 19 what would be the remainder?
Solution:
Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient
and 9 as remainder.
18.In doing a division of a question with zero remainder,a
candidate took 12 as divisor instead of 21.The quotient
obtained by him was 35. The correct quotient is?
a.0 b.12 c.13 d.20
Solution:
Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.
19.If a number is multiplied by 22 and the same number is
added to it then we get a number that is half the square
of that number. Find the number.
a.45 b.46 c.47 d. none
Solution:
Let the required number be x.
Given that x*22+x = 1/2 x2
23x = 1/2 x2
x = 2*23=46
Ans (b)
20.Find the number of zeros in the factorial of the number 18?
Solution:
18! contains 15 and 5,which combined with one even number
gives zeros. Also 10 is also contained in 18! which will
give additional zero .Hence 18! contains 3 zeros and the
last digit will always be zero.
21.The sum of three prime numbers is 100.If one of them
exceeds another by 36 then one of the numbers is?
a.7 b.29 c.41 d67.
Solution:
x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.
22.A number when divided by the sum of 555 and 445 gives
two times their difference as quotient and 30 as remainder .
The number is?
a.1220 b.1250 c.22030 d.220030.
Solution:
Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
23.The difference of 1025-7 and 1024+x is divisible by 3 for x=?
a.3 b.2 c.4 d.6
Solution:
The difference of 1025-7 and 1024+x is
=(1025-7)-(1024-x)
=1025-7-1024-x
=10.1024-7 -1024-x
=1024(10-1)-(7-x)
=1024*9-(7+x)
The above expression is divisible by 3 so we have to
replace x with 2.
Ans (b).
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<b>Complex Problems:</b>
24.Six bells commence tolling together and toll at intervals
of 2,4,6,8,10,12 seconds respectively. In 30 minutes how many
times do they toll together?
Solution:
To find the time that the bells will toll together we have
to take L.C.M of 2,4,6,8,10,12 is 120.
So,the bells will toll together after every 120 seconds
i e, 2 minutes
In 30 minutes they will toll together [30/2 +1]=16 times
25.The sum of two numbers is 15 and their geometric mean is
20% lower than their arithmetic mean. Find the numbers?
a.11,4 b.12,3 c.13,2 d.10,5
Solution:
Sum of the two numbers is a+b=15.
their A.M = a+b / 2 and G.M = (ab)1/2
Given G.M = 20% lower than A.M
=80/100 A.M
(ab)1/2=4/5 a+b/2 = 2*15/5= 6
(ab)1/2=6
ab=36 =>b=36/a
a+b=15
a+36/a=15
a2+36=15a
a2-15a+36=0
a2-3a-12a+36=0
a(a-3)-12(a-3)=0
a=12 or 3.
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