@shattereddream said:take the lcm of 1,2 ,3, 4,5,6,7 ie 420
Bhai sahab logic kya hai ?
How many natural nos from 1 to 100 can be the LCM of 2 or more distinct natural nos. in only one way (x,y = y,x) ?
@sunnychopra89 said:Bhai sahab logic kya hai ?
Logic is that P has to be LCM of Q+1,Q+2,...,Q+7 t make our condition true
So, for min value of P, Q needs to be 0.
So, for min value of P, Q needs to be 0.
@sunnychopra89 said:How many natural nos from 1 to 100 can be the LCM of 2 or more distinct natural nos. in only one way (x,y = y,x) ?
All prime numbers 25 ?
@sunnychopra89 said:[(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ............ + (1000!)^(1000!)] / [2^3 X 5^1].Find the Remainder.
Do seperately for 2^3 and 5..==> 8a+5 = 5b+7 ==> a=4 ==> rem =37
@The_Loser said:in hw many ways alphabet of letter be arranged so that dre are 7 letters bw a & b.
18*2 * 24c7 ??
@ravi.theja said:Do seperately for 2^3 and 5..==> 8a+5 = 5b+7 ==> a=4 ==> rem =37
Bhai nei palle padra yaar....CRT mujhe samaj hi nei aa raha saala...Can u plz brief it
Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?
In a 50 digit +ve no., all the digits are 4 except for the nth digit. This no. is divisible by 13 for some choice of that nth digit. How many possible values can n have?
@shattereddream said:Logic is that P has to be LCM of Q+1,Q+2,...,Q+7 t make our condition true So, for min value of P, Q needs to be 0.
But this way P+1 comes out to be 421 which is a prime. Where am I missing here? :/
@sunnychopra89 said:P and Q are 2 distinct whole nos. and P+1, P+2....P+7 are integral multiples of Q+1, Q+2....Q+7 respectivly. What is the min value of P?
LCM of 1,2,3,4,5,6,7 i.e. 420 i guess?
@shattereddream said:Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?
abcd-ab+cd=5481
ab*100+cd-ab+cd=5481
99ab+2cd=5481
cd has to be a multiple of 9
possible sol-55,18
73
@sunnychopra89 said:Dude Approach plzzz?
It's satisfying the property of prime no's. Try it out by substituting any of them.
@shattereddream said:Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?
Shouldn't the number be 5491 ?
THe sum of first 4 terms of ap is 28 and sum of first 8 term of the same ap is 88. Find the sum of first 16 terms of the ap?
@shattereddream said:Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?
55,18
5518-(55-18)=5481
55+18=73?
if a times the ath term of an ap is equal to the b times the bth term, find (a+b)th term?
1. 0 3. a^2 + b^2
2. a^2 - b^2 4. a - b
5. 1
@x2maverickc said:But this way P+1 comes out to be 421 which is a prime. Where am I missing here? :/
p+1 is a multiple of q+1 , p+2 is a multiple of q+2 & so on .....
keep q =0 (as small as possible for minimum)
p+1 should be divisible by 1
p+2 should be divisible by 2 , as 2 is divisible by 2 hence p has to be divisible by 2 similarly for others too which means p should be divisible by 1,2,3,4,5,6,7 ...... according to me