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1. Problems on numbers

2. 1. How many zeroes are there at the end of 100!?
2. Find the remainder when 1! + 2! + 3! …50! Is divided by 7?
3. The sum of 2 numbers is 156 and their HCF is 13. Find the number of pairs
of such numbers possible?
4. If A30158B is a 7 digit number, which is exactly divisible by 72. then find the
value of A?
5. 28 apple trees, 42 Mango trees, and 56 orange trees have to be planted in
rows such that the number of trees in each row are equal and of same
variety. Find the minimum number of rows required to plant the trees?
6. ABCDE 4 = EDCBA . If each letter represents a different digit then find the
value of A+B+C+D+E?

3. A man had some chocolates with him. If he distributes equally among 36
students, then 32 chocolates will remain with him. If he distributes equally
among 45 students, then 41 chocolates will remain with him. If he distributes
equally among 54 students, then 50 chocolates will remain with him. Find the
minimum number of chocolates , the man can have?
168 red color balls, 192 blue color balls, and 264 green color balls have to be
placed in boxes such that the number of balls in each box are equal and of
same color. Find the minimum number of boxes required to place the balls?
What is the max number of years, a person has to wait to celebrate his
birthday?
7.
8.
9.

4. How many zeroes are there at the end of 100! And 140! ?
Ans :
Prime Numbers – 2,3,5,7,11,13,17,19………….…73,79,83,87,97
Except 2,3 remaining prime numbers are of the form 6n 1
= 0
20 + 4 + 0 = 24 28 + 5 + 1 + 0 = 34
1.
Total number of zeroes at the end of 100! = 24 and 140! = 34

5. Find the remainder when 1! + 2! + 3! …50! Is divided by 7?2.
Ans :
6 + 4 = = 3
Similarly
1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! …… + 50!
1 2 6 3 1 6 0 0 ……… 0 = = 5
Remainder = 5
Remainder = 5
Remainder

6. The sum of 2 numbers is 156 and their HCF is 13. Find the number of
pairs of such numbers possible?
3.
Ans :
NOTE :
• Every number is a factor of itself. ‘1’ is a factor of every number.
• CO – PRIMES : if the HCF of 2 numbers is ‘1’ then they are co-primes.
• 2 even numbers can never be co-primes
From the given question A + B = 156 and HCF( A , B ) = 13
 A and B are multiples of 13. let A = 13x and B = 13y
 13x + 13y = 156
 x + y = 12 => Pairs possible are
 X,Y = 1,11 or 5,7
2 pairs are possible
1,11;
2,10;
3,9;
4,8;
5,7;
6,6

7. 2 – units place an even number should be present
3 – sum of digits should be divisible by 3
4 – last 2 digits should be divisible by 4 and 00 case
5 – last digit should be ‘5’ or ‘0’.
6 - last digit should be divisible by 2, 3
8 – last 3 digits should be divisible by 8 or 000 case
9 – sum of digits by divisible by 9
11 – (sum of alternatively digits – sum of other digits) should be zero or
should be divisible by 11

8. If A30158B is a 7 digit number, which is exactly divisible by 72. then
find the value of A?
4.
Value of A = 6
Ans :
72 can be expressed as 9 * 8
 The given number should be divisible by 9 and 8
 A+3+0+1+5+8+B = 17+A+B should be divisible by 9
 58B should be divisible by 8 => B = 4
 21+A should be divisible by 9 => A =6

9. 28 apple trees, 42 Mango trees, and 56 orange trees have to be
planted in rows such that the number of trees in each row are equal
and of same variety. Find the minimum number of rows required to
plant the trees?
5.
Ans :
Based upon the two conditions i.e. number of trees in each row should
be equal and of same variety .
Number of trees in each row can be 1,2,7,14 (common factors of 28,42,56)
We have to find the minimum number of rows
 HCF should be maximum. Therefore maximum number of trees in a row is 14
 number of rows are 28/14 + 42/14 + 56/14 = 9.
Number of rows are = 9

10. ABCDE 4 = EDCBA . If each letter represents a different digit then
find the value of A+B+C+D+E?
6.
Ans :
ABCDE 4 = EDCBA
 A = 2
 E = 8
 B = 1
 D = 7
 C = 9
A+B+C+D+E = 27
A – 0,2,4,6,8
BA - should be divisible by 4
B – 1,3,5,7,9
(D * 4 ) + 3 = 1 => D = 2, 7
C – 8 ,9

11. A man had some chocolates with him. If he distributes equally among 36
students, then 32 chocolates will remain with him. If he distributes equally
among 45 students, then 41 chocolates will remain with him. If he distributes
equally among 54 students, then 50 chocolates will remain with him. Find the
minimum number of chocolates, the man can have?
7.
Ans :
36) x ( 45) x ( 54) x (
___ ___ ___
32 41 50
+4 or -32 +4 or -41 +4 or -50
X+4 is the common multiple of 36, 45, 54
Therefore LCM of 36, 45, 54 = 540
=> X + 4 = 540 => x = 536.
Number of chocolates = 536

12. 168 red color balls, 192 blue color balls, and 264 green color balls have to be
placed in boxes such that the number of balls in each box are equal and of
same color. Find the minimum number of boxes required to place the balls?
8.
Ans :
Based upon the two conditions i.e. number of balls in each box should be
equal and of same color .
HCF of 168, 192, 264 = 24.
Number of boxes = 168/24 + 192/24 + 264/24 = 26
Maximum number of balls in each box = 24
 Number of boxes = 26
Number of Boxes = 26

13. What is the max number of years, a person has to wait to celebrate his
birthday?
9.
Ans :
A leap year should be divisible by 4. but if a leap year that is ending with 00
then that should be divisible by 400. then only it is called as Leap year.
1896 , 1900, 1904 …in this series 1900 is not a leap year
Therefore a person born on 29th February in leap year have to wait again
for 1904.
Maximum number of years is 8
Maximum number of years is 8

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Problems for practice are given in the description of the video.

15. Practice problems
on Numbers

16. 1. What is the power of 7 in 100!?
2. Find the remainder when 1! + 2! + 3! …50! Is divided by 5?
3. The sum of 2 numbers is 187 and their HCF is 17. Find the
number of pairs of such numbers possible?
4. The product of 2 numbers is 1728 and their HCF is 12. Find
the number of pairs of such numbers possible?

17. What is the power of 7 in 100! ?
Ans :
Prime Numbers – 2,3,5,7,11,13,17,19………….…73,79,83,87,97
Except 2,3 remaining prime numbers are of the form 6n 1
0
14 + 2 + 0 = 16
1.
Power of 7 in 100! Is 16.

18. Find the remainder when 1! + 2! + 3! …50! Is divided by 5?2.
Ans :
0
Similarly
1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! …… + 50!
1 2 6 24 0 0 0 0 ……… 0 = = 3
Remainder = 3
Remainder = 3

19. The sum of 2 numbers is 187 and their HCF is 17. Find the number of
pairs of such numbers possible?
3.
Ans :
NOTE :
• Every number is a factor of itself. ‘1’ is a factor of every number.
• CO – PRIMES : if the HCF of 2 numbers is ‘1’ then they are co-primes.
• 2 even numbers can never be co-primes
From the given question A + B = 187 and HCF( A , B ) = 17
 A and B are multiples of 17. let A = 17x and B = 17y
 17x + 17y = 156
 x + y = 11 => Pairs possible are
 X,Y = 1,10 ; 2,9; 3,8; 4,7; 5,6
5 pairs are possible
1,10;
2,9;
3,8
4,7
5,6;

20. The product of 2 numbers is 1728 and their HCF is 12. Find the number of
pairs of such numbers possible?
3.
Ans :
NOTE :
• Every number is a factor of itself. ‘1’ is a factor of every number.
• CO – PRIMES : if the HCF of 2 numbers is ‘1’ then they are co-primes.
• 2 even numbers can never be co-primes
From the given question A * B = 1728 and HCF( A , B ) = 12
 A and B are multiples of 12. let A = 12x and B = 12y
 12x * 12y = 1728
 x + y = 12 => Pairs possible are
 X,Y = 1,12 ; 4,3
2 pairs are possible
1,12;
4,3
2,6;

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