If you read Number System in Quantitative section. Tutorial Diary presents Elitmus Quantitative Ability - Number System important and previous questions with answers and solutions.

Elitmus Number System Questions-1 | Elitmus Papers

**Question:**What is the remainder of 18!/23?

(a) 1

(b) 2

(c) 3

(d) 4

**Answer: (a)**

**Solution:**

Let rem[18!/23]=r

Sincw, Wilson's Theorem says, rem[(p-2)!/p]=1

rem[21!/23]=1

rem[21*20*19*18!/23]=1

rem[(-2)(-3)(-4)*r/23]=1

rem[-24r/23]=1

24r=-23K+1 ... k=-1

r=23+1/24

r=1

Therefore, Answer is (a).

**Question:**If N= 1!+2!+3!.....+10!. Then, what will be the last digit of N^N?

(a) 3

(b) 7

(c) 1

(d) 4

**Answer: (b)**

**Solution:**

1! + 2! + 3! + 4! + 5! + 6! +7! +8! +9! +10!

1 + 2 + 6 + 24 + 120+........(from here all the last digits are zeros)................

The last digit of above sum is 3

So if we cube 3 ==> (3^3)=27

So the answer is 7

**Question:**A set M contains element all even number between 1 and 23 and all odd numbers 24 and 100. if all the elements of the set multiplied than how many trailing 0 , resulting number will contains?

(a) 10

(b) 12

(c) 9

(d) 13

**Answer: (b)**

**Solution:**

m=(2,4,6,8,10,12,14,16,18,20,22,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,87,89,91,93,95,97,99)

question asked how many trailing 0

there are two condition

1st if we multiply 2*5=10

end with =0

if we count 5 in set m we get 12

if we count 2 in set m we get more than 12

so total no end with 0 equal to 12

**Question:**if p is number greater than 99 ,e is the probability of divisiblity of any even whole number p, and O is the probablity of any odd whole number divisible by p , w is probablity of any whole number divisible by p.

(a) w=2e

(b) e=w/2

(c) o=w/2

(d) either (b) or (c)

**Answer: (d)**

**Question:**how many no. will be possible between 100 to 500 such that the sum of two digit is equal to the third(example -312, 2+1=3)??

a)12

b)24

c)66

d)72

**Answer:**(c)

**Solution:**for 123 total permutation 123,132,213,231,312,321

123=6

134=6

145=4 because in 6 case two case 541>500 and 514>500

156=2 because in 6 case four case 651>500 and 615>500 and 516 ,561

167=2

178=2

189=2

235=4

246=4

257=2

268=2

279=2

347=4

358=2

369=2

total=46

and

101,110

202,220

303,330

404,440

total=8

121,221,112

242,224,422

336,363

448,484

total=10

Total no of case 46+8+10=66

**Question:**find remainder of (9^1+9^2+.........+9^n)/6

n is multiple of 11.

(a) 0

(b) 5

(c) 3

(d) can not be determine

**Answer: (c)**

**Solution:**

Since:

9^(1+2+3+11n)/6

=>9^(11n(11n+1)/2)/6 (suppose it is 11)

=>9^11(11+1)/6*2=3(ans)

=>9^22(22+1)/6*2=3(ans)

**Question:**the square of two digit number(ab) divided by half of itself resultant added by 18 then divided by 2 then we get reverse of original number(ba). how many combination of ab exist

(a) 9

(b) 8

(c) 7

(d) 6

**Answer: (b)**

**Solution:**

(((10a+b)^2)/(10a+b/2)+18)/2=10b+a.

Simplify the above equation.

The final equation will be a-b=-1

There will be 8 combinations of a &b which will satisfy this equation.

i.e.

a=1,b=2

a=2,b=3

a=3,b=4

a=4,b=5

a=5,b=6

a=6,b=7

a=7,b=8

a=8,b=9

**Question:**Find the 100th place digit of a the greatest number forming from digits 1 to 9 without repeatation and number passes the divisibility test of 11 means difference of sum of alternative digits is 0 or divisible by 11.

(a) 1

(b) 2

(c) 3

(d) 4

**Answer:**(d)

**Solution:**The 100th place digit would be 4.

Since the largest number which is divisible by 11 having digits 1 to 9 is 978653412,

the difference of the sum of the digits at even place and odd place is 11, which is a multiple of 11.

Hope you like this, share it with your friends to help them.