Problems on numbers
Posted by
Ravi Kumar at Monday, November 24, 2008
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Problems on numbers
Problems on numbers
1.Simplify
a.8888+888+88+8
b.119927823456
Solution: a.8888
888
88
8
9872
b.119927823456=11992(7823+456)
=119928279=3713
2.What could be the maximum value of Q in the following equation?
5PQ+3R7+2Q8=1114
Solution: 5 P Q
3 R 7
2 Q 8
11 1 4
2+P+Q+R=11
Maximum value of Q =112=9 (P=0,R=0)
3.Simplify: a.5793405*9999 b.839478*625
Solution:
a. 5793405*9999=5793405*(100001)
579340500005793405=57928256595
b. 839478*625=839478*54=8394780000/16=524673750.
4.Evaluate 313*313+287*287
Solution:
a²+b²=1/2((a+b)²+(ab)²)
1/2(313+287)² +(313287)²=1/2(600 ² +26 ² )
½(360000+676)=180338
5.Which of the following is a prime number?
a.241 b.337 c.391
Solution:
a.241
16>√241.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
241 is not divisible by any of these. Hence we can
conclude that 241 is a prime number.
b. 337
19>√337.Hence take the value of Z=19.
Prime numbers less than 16 are 2,3,5,7,11,13 and 17.
337 is not divisible by any of these. Hence we can conclude
that 337 is a prime number.
c. 391
20>√391.Hence take the value of Z=20.
Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.
391 is divisible by 17. Hence we can conclude
that 391 is not a prime number.
6.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.
7.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
8.Which of the following numbers s divisible by 3?
a.541326
b.5967013
Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.
b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.
9.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.
10.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 = 1917=2.
11.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 =111100=11.
Hence,required number =10011.
12.On dividing 15968 by a certain number the quotient is 89 and the remainder
is 37.Find the divisor?
Solution:Divisor = (DividendRemainder)/Quotient
=(1596837) / 89
=179.
13.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.
14.A number being successively divided by 3,5,8 leaves remainders 1,4,7
respectively. Find the respective remainders if the order of
divisors are reversed?
Solution:Let the number be x.
3 x 5 y  1 8 z  4 1  7 z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now
8 238
5 29  6
3 5  4
1  2
Respective remainders are 6,4,2.
15.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.
16.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+(n1)d
=14+(n1)7
=7+7n
7n=77 =>n=11.
17.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+(n1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.
18.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 n1=1024
2n1 =512=29
n1=9
n=10.
19.2+22+23+24+25..........+28=?
Solution:Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1r n)/1r=Sn=2(128)/12.
=2*255=510.
20.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.
21.The sum of all possible two digit numbers formed from three
different one digit natural numbers when divided by the sum of the
original three numbers is equal to?
a.18 b.22 c.36 d. none
Solution:Let the one digit numbers x,y,z
Sum of all possible two digit numbers=
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)
= 22(x+y+z)
Therefore sum of all possible two digit numbers when divided by sum of
one digit numbers gives 22.
22.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.
23.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)*(555445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
24.The difference between two numbers s 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.
25.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.