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
The number is 34369 - 31513 = 2856 can divide both 34369 and 31513 leaving same remainder 97, hence ans is b) 97 3 digit nos are 2856/3 = 952, 2856/4 = 714, 2856/6 = 476, so on...
The number is 34369 - 31513 = 2856 can divide both 34369 and 31513 leaving same remainder 97, hence ans is b) 97 3 digit nos are 2856/3 = 952, 2856/4 = 714, 2856/6 = 476, so on...
hey can u xpln why this 34369-31513 will be tht no. can u xpln the logic behind it ??? btw thnx fr this rply Sahana vry cnfsng name
hey can u xpln why this 34369-31513 will be tht no. can u xpln the logic behind it ??? btw thnx fr this rply Sahana vry cnfsng name :biggrin:
Just check it with smaller numbers, so as to get the logic by yourself, ex 7 and 10, difference is 3, which leaves 1 as remainder. 11 and 7, difference is 4, which leaves 3 as remainder.
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
The number is 34369 - 31513 = 2856 can divide both 34369 and 31513 leaving same remainder 97, hence ans is b) 97 3 digit nos are 2856/3 = 952, 2856/4 = 714, 2856/6 = 476, so on...
asd_dah Says
hey can u xpln why this 34369-31513 will be tht no. can u xpln the logic behind it ??? btw thnx fr this rply Sahana vry cnfsng name :biggrin:
Let the three digit no. be N then according to question.....
34369 = k1*N + R 31513 = k2*N + R.............subtracting these two eqns......
2856 = (k1-k2)N...........
so N is a factor of 2856 (Thats why we subtract)
But when we start factorising 2856 we get 2856 = 2*2*714 or second combo could be 2*2*2*357 or the subsequent ones upto 119, we can have solutions from all those three digit no.
Hence one of the possible value of N is 714.......... Hence we can find the remainder now, which comes out to be 97
Slightly twisted from the previous post, enjoy solvin.. Let S = (3 + 3^2 + 3^3 + + 3^800) (7 + 7^2 + 7^3 + + 7^401). The last three digits of S are a.143 b. 993 c. 003 d. 907
Why did you not consider 78? and 69? and 87? and 96?
Beats me!
u have to check for d condition that 9(x-y)/(x+y) is an integer, but if u take x as 8 u get y as 1,4 and y as 7 does not satisfy d condition so u have to check it..... as x+y should be a multiple of 3 is a necessary but not sufficient condition........
Slightly twisted from the previous post, enjoy solvin.. Let S = (3 + 3^2 + 3^3 + + 3^800) - (7 + 7^2 + 7^3 + + 7^401). The last three digits of S are a.143 b. 993 c. 003 d. 907
Originally Posted by Sahana Kavitha View Post Slightly twisted from the previous post, enjoy solvin.. Let S = (3 + 3^2 + 3^3 + + 3^800) - (7 + 7^2 + 7^3 + + 7^401). The last three digits of S are a.143 b. 993 c. 003 d. 907
Getting (b) 993
As S = /2 - /6
Now (3^801 -3) mod 1000 = 000 and (7^402 -7) mod 1000 = 042
S = (XXX...000)/2 - (XXX....042)/6
Now (3^801 -3) is divisible by 2 so 2a = 1000b 1000 and also
(7^402 -7) is divisible by 6 so 6a = 1000b + 42 3042
Now (3^801 -3) mod 1000 = 000 and (7^402 -7) mod 1000 = 042
S = (XXX...000)/2 - (XXX....042)/6
Now (3^801 -3) is divisible by 2 so 2a = 1000b 1000 and also
(7^402 -7) is divisible by 6 so 6a = 1000b + 42 3042
Now
S = (XXX...1000)/2 - (XXX....3042)/6
S = XXX....500 - XXX....507
S = XXX....993
dude the last three steps are a bit confusing...u arrive at 000 and 042 after dividing them by 2 and 6 respectively isn't it? or is it tht i m mistaken...pls clear my doubt:)
dude the last three steps are a bit confusing...u arrive at 000 and 042 after dividing them by 2 and 6 respectively isn't it? or is it tht i m mistaken...pls clear my doubt:)
Your method is perfect......... i.e. now we have the last 4 digits as 1000 and 3042 now eventually we have to divide them by 2 and 6 respectively.......... So when 1000 divided by 2 we have 500 as last three digits and when 3042 is divided by 6 we have 507 as last three digits, irrespective of the digits preceding 1 and 3 .........
1) x>y>z>=2 and xyz= 2002 where x,y,z are natural numbers. So whats the Max. and Min. Value of x+y+z ?
hi, xyz= 2002 can be written as xyz=2*7*11*13 minimum value occurs when x,y,z are close to each other in value... i.e. 11 13 14 which becomes...38 max will be 143+7+2 = 152 hope it is correct...
answer is 1