Find the maximum value of n such that 50! is perfectly divisible by 12600 raised to the power n.
10?
12600= (2^3)* (3^2)*(5^2)*(7)
powers of 2 in 50! by successive division = 50+25+12+6+3+1= 97, hence power of 2^3 will be 97/3 = 32
same way powers of 3 = 16+5+1 = 22; powers of 3^2 = 11
powers of 5 = 10+2 = 12; powers of 5^2 = 6
powers of 7 = 7+1 = 8
from the above set since there are only 6 powers of 5^2 are available, hence maximum value of n can be 6 only
Common factors means Factors of HCF ( P,Q)
HCF (P,Q) = 27 x 35 x 73
For perfect squares , we can select :
20, 22 , 24 , 26 in 4 ways
30 , 32 , 34 in 3 ways
70 , 72 in 2 ways
We need even perfect square so at least one 2 should be there
Hence , 3 x 3 x 2 = 18 factors ..
Let the three consecutive odd integers be a-2 , a and a + 2
(a -2)2 + a2 + (a +2)2 = 3a2 + 8
Only option C gives remainder 2 when divided by 3 Hence , 5555.
How many 18-digit positive integers are there which ends in 18 as last two digits and are divisible by 18?
18 = 2 x 9 so the number has to be divisible by 2 and 9 both.
From the question it is clear that the number ends with 18 , hence it is divisible by 2 .
Sum of remaining digits must be a multiple of 9 .
Hence all 16 digit numbers divisible by 9 should be the answer.
From 1015 to 1016 -1 there are [1016 – 1015]/9 = 1015 multiples of 9.
| A and B when divided by 56 leave remainders of 48 and 32 respectively. When divided by 44, both leave a remainder of 24. If the sum of A and B is divisible by the sum of the divisors, then find the minimum value of (A + B). |
A=56a+ 48
B=56b+ 32
A+B= {56(a+b) +80}
A and B leave reminder 24 when divided by 44
A+B will leave reminder: (24+24)/44 = 4
A+B={ 56(a+b)}/44= 4(reminder)
Check value of a+b by putting 1,2,3.....
a+b =1 satisfies here and it is minimum hence
A+B= 56*(1)+80
=136 (answer)
Nine distinct digits appear in the product of 2, 7,181, 241 and 607. Which digit is missing?
(a)1 (b)2 (c)4 (d)6
The digital sum of a number N is same as the remainder obtained by dividing the number N by 9.
Digital sum of the product 2 x 7 x 181 x 241 x 607 is 2 x 7 x 1 x 7 x 4 = 28 x 14 => 5 x 1 ;i.e 5.
Since the sum of all digits from 0 to 9 (1234567890) is 45, which is divisible by 9.
The only digit missing that could produce a remainder of 5 by 9 when divided by 9 is 4.
How many natural numbers from (111)6 to (1111)6 are divisible by 9?
Decimal representation of 1116 = 62 x 1 + 6 x 1 + 1 = 43
Decimal representation of 11116 = 63x1 + 62x 1 + 6 x 1+ 1= 259 .
Now , count multiples of 9 between these two numbers : [ 252 - 45]/2 + 1 = 24 Numbers .
Let the product of first thousand even positive integers is A and product of first thousand odd positive integers is B. Find the remainder when A – B is divided by 2001?
Product of first 1000 even numbers : 2 x 4 x 6 x 8 x........x 2000
= 2^1000 x ( 1 x 2 x 3 x ..... x 665 x 666 x 667 x ........ x 999 x 1000) = A
Product of first 1000 odd numbers : 1 x 3 x 5 x 7x .... 665 x 666 x 667 ... x 1999 = B
2001 = 3 x 667
A and B both the numbers are divisible by 3 and 667 .
hence net remainder 0.
How many of the following numbers are prime?
I. 172012 – 19
II. 152013 – 13
III. 142031 – 27
I. 172012 – 19 => difference of two odd number is always even so composite
II. 152013 – 13 => even , divisible by 2 Hence , composite
III. 142031 – 27 => (14677 )3 – 33
( A3 - B3) is always divisible by (A - B), so 142031 – 27 is a composite number.
Hence , None
if n=539*2^18 and m=9*2^13 , then the remainder when n is divided by m is?
Hi sir, please guide
{(1000+729)^752}/144*12
{(1729)^752}/1728
Remainder=1
77777...... 1001 times
Any digit repeated 6 times is always divisible by 1001 so
77777.... Written 996 times is a multiple of 1001
Hence , 77777 /1001=>
777 -77 = 700
Take 1075^n common
You will get 1075^(n-1) *1074
Now check options 15 will be divisible because 1075 contains 5 and 1074 contains 3
1075 contains 43
1074 contains 179 we can find it by prime factorisation
Hence 23 is not divisible
Since digits of 10digits number n sum to 45 , n should be divisible by 9
n should be divisible by 99999
10 digits number will be of type
n = 100000x + y where x is 5digit number and y is 4 or 5 digit number
n = 99999x + (x+y)
means x+y should be divisible by 99999
and because neither x > 99999 nor y>99999
x+y = 99999
x=abcde
y=pqrst
----------------------
x+y=99999
also since max numbers is 9 and 8 we cannot carry to next digit (19)
so it has to be all 9s without carry
groups
0 9
1 8
2 7
3 6
4 5
means combination is for x only. y is related
digit 1 : {1-9} 9ways
digit 2 : {digit1,0} - {digit1,9-digit1} 8 ways
digit 3 : {digit2} - {digit2.9-digit2} 6 ways
digit 4 : {digit3} - {digit3.9-digit3} 4 ways
digit 5 : {digit4} - {digit4.9-digit4} 2 ways
total ways : 9 × 8 × 6 × 4 × 2 =3456
from 1sr statement there are two possibilities...2^2*3^6 or 2^5*3^3
(considering that from 2nd statement on multiplying n with 3 or n with 2 , factors don't double or get hugely increased hence 2 and 3 are the only prime factors)
now from 2nd statement we find that 1st case in not possible hence, our n= 2^5*3^3
now u can find 6n, (6+1)*(4+1) =35
25 pages consist of 25 odd numbers and 25 even numbers
So there sum should always be odd
Hence 1900 an even sum is never possible.
