Q. How many two digit numbers less than or equal to 50, have the product of the factorial of their digits less than or equal to the sum of factorials of their digits?
I got 18; however the answer is 17. Please explain.
Thanks,
Rohit.
Hi Guys,
Could you please explain how to approach following two types of problems?
1. Find last two digits of 65*29*37*63*71*87
65*29*37*63*71*87 mod 100
13*29*37*63*71*87 mod 20
(-7)*9*(-3)*(3)*11*7 mod 20
59 mod 20 = 19
Answer => 19*5 = 95
Hi Guys,
Could you please explain how to approach following two types of problems?
1. Find last two digits of 65*29*37*63*71*87
2. What is the greatest power of 12 in 123!
Thanks,
Rohit.
What is the greatest power of 12 in 123!
12=2^2*3
so
123/2 = 61
123/4 = 30
123/8 = 15
123/16 = 7
123/32 = 3
123/64 = 1
61 + 30 + 15 + 7 + 3 + 1 = 117
117/2 = 58
For 3,
123/3=41
123/9=13
123/27=4
123/81=1
so greatest power of 12 in 123!= 58
2. What is the greatest power of 12 in 123!
Thanks,
Rohit.
12 = 2^2 * 3
power of 2^2
123/2 = 61
123/4 = 30
123/8 = 15
123/16 = 7
123/32 = 3
123/64 = 1
so total = 61 + 30 + 15 + 7 + 3 + 1 = 117
117/2 = 58
power of 3
123/3 = 41
123/9 = 13
123/27 = 4
123/81 = 1
so totally 41 + 13 + 4 + 1 = 59
so answer is min(58,59) = 58
Q. How many two digit numbers less than or equal to 50, have the product of the factorial of their digits less than or equal to the sum of factorials of their digits?
I got 18; however the answer is 17. Please explain.
Thanks,
Rohit.
Let the no. be "ab" then according to question.........
a! * b!
Now for it to be less or equal to the sum we have.......
11 to 19 - 9 values..........
21 , 31 , 41 - 3 values..........
10, 20, 30, 40, 50 - 5 values.........
22 - 1 value
So a total = 9 + 3 + 5 = 18 values.......
Thanks Rohit (Edited)..........
From previous post,
1. reminder for 10^1729%1729?
2. last 3 digits of 7^64
From previous post,
1. reminder for 10^1729%1729?
1729 = 7*247 = 7*13*19 = 91*19
10^1729 = 10 mod 91
10^1729 = 10 mod 19
So final answer is 10.
From previous post,
2. last 3 digits of 7^64
7^64 mod 1000
7^64 mod 8 = 1
7^64 mod 125
7^4 mod 125 = 26
(7^4)^16 mod 125 = 26
125a+26 = 8b+1
Put a=3
Last Three Digits = 401
Let the no. be "ab" then according to question.........
a! * b!
Now for it less then the sum, one of the digits must be 1 so we have.......
11 to 19 - 9 values..........
21 , 31 , 41 - 3 values..........
for it to be equal to sum, one of the digits must be 0 so we have .........
10, 20, 30, 40, 50 - 5 values.........
So a total = 9 + 3 + 5 = 17 values.......
Shouldn't 22 be included as well?
Rohit.
65*29*37*63*71*87 mod 100
13*29*37*63*71*87 mod 20
(-7)*9*(-3)*(3)*11*7 mod 20
59 mod 20 = 19
Answer => 19*5 = 95
Thanks for the response. Could you please elaborate on the procedure?
Rohit.
1. 32^32^32 / 7 will leave a remainder of:
(a) 4
(b) 7
(c) 1
(d) 2
2. Suppose the product of n consecutive integers is x.
(x+1)(x+2)...(x+(n-1)) = 1000, then the number of terms n can be
(a) 16
(b) 5
(c) 25
(d) 20
3. How many integer values of x and y satisfy the expression 4x+7y=3 where x
Kindly provide an explanation too.
Thanks,
Rohit.
1. 32^32^32 / 7 will leave a remainder of:
(a) 4
(b) 7
(c) 1
(d) 2
3. How many integer values of x and y satisfy the expression 4x+7y=3 where x
Kindly provide an explanation too.
Thanks,
Rohit.
(1). Getting (a) 4
We can apply Euler's
E(7) = 6
now 32^32 mod 6 or
(2^5)^32 mod 6 or
2^160 mod 6 or
2^159 mod 3 = 2
So 2^160 mod 6 = 2*2 = 4
So
32^32^32 mod 7 = 32^(6k + 4) mod 7 = 32^4 mod 7 = 4
(3). Getting 286 values
4x+7y=3 first value which satisfies :- x = -1 , y = 1 (for x = -ve and y +ve as both can't be -ve or +ve)
Now 'x' will jump with a factor of -7 and 'y' will jump with a factor of 4
So next value x = -8 and y = 5 and so on
So values of x will fall early under 1000 (-1,-8,-15........upto 1000)
It is an AP no. of terms = 143
Now for x = +ve and y = -ve first value x = 6 and y = -3
Hereon 'x' will jump with a factor of 7 and 'y' with a factor of -4 so next value x = 13 and y = -7 and so on
x = 6,13,20 ....upto 1000) No. of terms = 143
So total 286
1. 32^32^32 / 7 will leave a remainder of:
(a) 4
(b) 7
(c) 1
(d) 2
2. Suppose the product of n consecutive integers is x.
(x+1)(x+2)...(x+(n-1)) = 1000, then the number of terms n can be
(a) 16
(b) 5
(c) 25
(d) 20
3. How many integer values of x and y satisfy the expression 4x+7y=3 where x
Kindly provide an explanation too.
Thanks,
Rohit.
1. (a) 4
2.b) 5
(x+1),(x+2)...(x+(n-1)) are n-1 consecutive integers. product of n-1 consecutive integers should be divided by n-1
in options n-1 : 15,4,24,19
only 4 divides 1000
however product of 4 consecutive integers should be divisible by 3, so answer should be none of these
3. 286
1. (a) 4
2.b) 5
(x+1),(x+2)...(x+(n-1)) are n-1 consecutive integers. product of n-1 consecutive integers should be divided by n-1
in options n-1 : 15,4,24,19
only 4 divides 1000 so n=5
3. 286
For 2nd ques.......
(x+1)(x+2)...(x+(n-1)) = 1000
Now if the ans is 5 then we will have four consecutive nos. whose product will be 1000 as (x+1)(x+2)(x+3)(x+4) = 1000
Now these are four consecutive integers hence one would be a multiple of 3 but 1000 = 2*2*2*5*5*5.........
So there is no possible combination .........
Correct me if I am wrong....
For 2nd ques.......
(x+1)(x+2)...(x+(n-1)) = 1000
Now if the ans is 5 then we will have four consecutive nos. whose product will be 1000 as (x+1)(x+2)(x+3)(x+4) = 1000
Now these are four consecutive integers hence one would be a multiple of 3 but 1000 = 2*2*2*5*5*5.........
So there is no possible combination .........
Correct me if I am wrong....
yup, you are right, answer should be none of these
Hi PUYs, please help me out with this question!
A two-digit number having distinct digits when divided by the sum of the digits gives the same remainder as when a two-digit number that is formed by reversing the digits of original number is divided by the sum of the digits.How many such two-digit numbers are possible?
1. 12
2. 14
3. 16
4. 18
5. null
Hi PUYs, please help me out with this question!
A two-digit number having distinct digits when divided by the sum of the digits gives the same remainder as when a two-digit number that is formed by reversing the digits of original number is divided by the sum of the digits.How many such two-digit numbers are possible?
1. 12
2. 14
3. 16
4. 18
5. null
N=10x +y = (x+y)K + r
10y+x = (x+y)n + r
9(x-y) = (x+y)(k-n)
so 9(x-y)/(x+y) shud be an integer
for x = 0 we have y from 1 to 9
for x = 1 we have y = 0,2,5,8
for x=2 we have y = 0,1,4,7
for x=3 we have y = 0,6
i guess there shud be 18 such nos. wats the oa?
N=10x +y = (x+y)K + r
10y+x = (x+y)n + r
9(x-y) = (x+y)(k-n)
so 9(x-y)/(x+y) shud be an integer
for x = 0 we have y from 1 to 9
for x = 1 we have y = 0,2,5,8
for x=2 we have y = 0,1,4,7
for x=3 we have y = 0,6
i guess there shud be 18 such nos. wats the oa?
same approach as yours but x=0 cases are not possible(since two digit no's).similarly,10,20,30 are not possible because if we reverse them we do not get a two digit no.
still this approach will be time consuming.. finding all no's for x=1 to 9.
waiting for a shorter approach.
cannot guess the ans.
Hi PUYs, please help me out with this question!
A two-digit number having distinct digits when divided by the sum of the digits gives the same remainder as when a two-digit number that is formed by reversing the digits of original number is divided by the sum of the digits.How many such two-digit numbers are possible?
1. 12
2. 14
3. 16
4. 18
5. null
N=10x +y = (x+y)K + r
10y+x = (x+y)n + r
9(x-y) = (x+y)(k-n)
so 9(x-y)/(x+y) shud be an integer
for x = 0 we have y from 1 to 9
for x = 1 we have y = 0,2,5,8
for x=2 we have y = 0,1,4,7
for x=3 we have y = 0,6
i guess there shud be 18 such nos. wats the oa?
same approach as yours but x=0 cases are not possible(since two digit no's).similarly,10,20,30 are not possible because if we reverse them we do not get a two digit no.
still this approach will be time consuming.. finding all no's for x=1 to 9.
waiting for a shorter approach.
cannot guess the ans.
check the multiples of 3, as a 9(x-y)/(x+y) is an integer, x+y shall be a multiple of 3
12, 15, 18, 21, 24, 27, 36, 42, 45, 48, 51, 54, 63, 72, 81, 84
hence the ans is 16, any observations pls let me know
PLZ HELP me out puyz.....
When 34369 and 31513 are divided by a certain three digit number, the remainders are equal. Find the remainder.
a>79
b>97
c>87
d>78
e>Cannot be determined
a detailed soln. will be appreciated