in how many ways 600 can be expressed as the sum of two or more consecutive integers.... !!
please anyone explain this question its recent aimcat question
@[464164:sameersapre23] 5 is the answer?
@[295195:suhansarkar]
(2^1000) * 2 = (2^10)^100 * 2
Last 2 digit of 2^10 is 24.
(24)^even gives 76 as last 2 digits.
so 76*2 = 152.
as we need only last 2 digits
the ans is 52.
@[322978:orangejawan]
8...
32^32^32/9 = 5^32^32 = (5^3)^29^32/9 = (-1)^29^32 = (-1)^odd = -1
-1+9=8.
@[301957:sandestar] is it 18081
@Sg231 said: What is the remainder when 21^3+23^3+25^3+27^3 divided by 96? 1) 0 2) 72 3) 12 4) 48 5) 24
1) 0 is the answer??
using : (24+1)^3 +(24-1)^3 +(24+3)^3+(24-3)^3
arun sharma...lod-2 ques.30:
find number of zeros in 1^1*2^2*3^3*...*100^100.
i get no. of 0's as 650 and no. of 5's as 500.
answer is 1300...plz help..i m stuck
find number of zeros in 1^1*2^2*3^3*...*100^100.
i get no. of 0's as 650 and no. of 5's as 500.
answer is 1300...plz help..i m stuck
@taruniim said:arun sharma...lod-2 ques.30: find number of zeros in 1^1*2^2*3^3*...*100^100. i get no. of 0's as 650 and no. of 5's as 500. answer is 1300...plz help..i m stuck
Hi,
Probably you are missing to count 5s twice in case of 25,50,75,100
then no. of 5s = (105 x 10) + 25 + 50 +75 +100 =1300
then no. of 5s = (105 x 10) + 25 + 50 +75 +100 =1300...why (105*10)??u cnt only take 5 into account....am i right??u would have to take no. of zeroes too..
@taruniim said:then no. of 5s = (105 x 10) + 25 + 50 +75 +100 =1300...why (105*10)??u cnt only take 5 into account....am i right??u would have to take no. of zeroes too..
Hi,
No. of zeroes = No. of 5s as when a 2 combines with a 5 it gives one zero..
Now check the expression 2 is in abundant so check for no. of 5s
Eventually no. of 5s = no. of 0s
Now write those where 5 is present 5 10 15 20..........85 90 95 100..considering each contain one 5 so no. of 5s in 15^15 is 15
5+100, 10+95, 90+15... =105 ten times so 105 x 10= 1050
Now for extra 5s in 75 eg=> 75 =5 x 5 x 3.. there will be 75 no. of more 5s
So ans. 1050+25+75+100
Just remember in 100^100 there are 200 5s but in 90^90 there are 90 5s
thnx leo...
hi puys...
pls help with this question :
A rational number A/B , where A & B are co-prime, is converted into a decimal number. If both A & B are less than 100, then for how many values of B will A/B always be a terminating decimal ?
@RPriyankhaa said: hi puys...pls help with this question : A rational number A/B , where A & B are co-prime, is converted into a decimal number. If both A & B are less than 100, then for how many values of B will A/B always be a terminating decimal ?
Any number of the form of X/2^a*5^b ends with a terminating decimal.
1,2,4,8,16,32,64,
5,25,
10,20,40,50,80
so 14 values
1,2,4,8,16,32,64,
5,25,
10,20,40,50,80
so 14 values
@unregisterjyoti said: Q (1):There exists a 10 digit number such that the 1st digit from left represents the number of 0's in the number, the 2nd digit from left represents the number of 1's occurring in the number and so on until the 10th digit represents the number of 9's in the number. 1. The sum of the digits of the number is: 1. 8 2. 9 3. 10 4. 19 2. The number of 0s in the number is: 1. 9 2. 8 3. 7 4. 6 3. The number of 1s occurring in the number is: 1. 0 2. 1 3. 2 4. 3 Q(2)Rahul intends to draw one rectangle of integer sides with a pencil which can last for a maximum possible length of 100 units only. Let R denote the set of all possible distinct rectangles from which rahul can choose to draw one such rectangle. 1. The number of rectangles in set R is 1. 636 2. 601 3. 613 4. 625 2. All the rectangles in R are formed in to groups such that all the rectangles of same perimeter are in the same group. What is maximum number of groups that is possible? 1. 99 2. 96 3. 49 4. 51
i guess the no. must be 8100000000 , any other no. won't satisfy the condition
hence sum of digits= (2). 9
no. of 0's = (2) 8
no of 1's = (2) 1
am i right??
Find the number of three-digit numbers that can be expressed as the sum of four consecutive,five consecutive and seven consecutive positive integers.
1)9 2)10 3)11 4)14 5)6
pls show detaied sol
What is the remainder when 128^1000 is divided by 153 ?
a) 103
b) 145
c) 118
d) 52
plz show me the way how to do it. Thank you
Somebody please show me the approach of this question
@varunvarma91 said: What is the remainder when 128^1000 is divided by 153 ?a) 103b) 145c) 118d) 52plz show me the way how to do it. Thank you
@[588873:varunvarma91]
is 52 the answer ?
i have done it like this :
153 = 17*9
so find the remainder separately with 17 and 9
so we will get :
with 17 remainder =1
with 9 remainder = 7
so such a number on investigation is 52!