Find the remainder when 54^124 is divided by 17?
a. 4 b. 5 c. 13 d. 15
(If anyone knows shorter method, please share it)
@Tina_angel :
1] _ _ _ _ : the first gap can have 5 nos [1,2,3,4,5]. the next can also have 5 [all but first gap's]. next: 4 [all but first two]. last: 3 [all but first 3] = 5*5*4*3 = 300
2] Plates with 1 digit = 5
Plates with 2 digits = 5C2 *2! = 20
3 digits = 5C3*3! = 60
4 digits = 5C4*4! = 120
5 digits = 5! = 120.
Total plates : 5+20+60+120+120 = 325.
@Tina_angel : It'll be 360 if they consider starting a number with zero : 6*5*4*3*2 = 360 - but numbers don't start with zero - may be as nothing is specified, we'll have to consider zero as a starter .. 
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What will be the remainder when 10^10 + 10^100 + 10^1000 + ....10^10000000000 is
divided by 7
@Rshmi05
10^10/7 = (-4)^10/7 = (4^3)^3 * 4 /7 => Rem. = 4
10^100/7 = (4^33)^3 * 4 /7 = Rem. = 4
Similarly for others, so 4 + 4 + 4 + ..... 10 times /7 = 40/7 = Rem. = 5
Similarly for others, so 4 + 4 + 4 + ..... 10 times /7 = 40/7 = Rem. = 5
@Rshmi05 said:What will be the remainder when 10^10 + 10^100 + 10^1000 + ....10^10000000000 isdivided by 7
(7+3)^10/7 + (7+3)^100/7 .....(7+3)^10000000000/7
3^10/7 + 3^100/7 ....
((28-1)^3)*3/7 + ((28-1)^33)*3/7 ...
-1*3/7 + (-1*3)/7 .... 10 times
-1*30/7 = -2/7 = 5 rem
3^10/7 + 3^100/7 ....
((28-1)^3)*3/7 + ((28-1)^33)*3/7 ...
-1*3/7 + (-1*3)/7 .... 10 times
-1*30/7 = -2/7 = 5 rem
@sonic_boom said:(7+3)^10/7 + (7+3)^100/7 .....(7+3)^10000000000/73^10/7 + 3^100/7 ....((28-1)^3)*3/7 + ((28-1)^33)*3/7 ...-1*3/7 + (-1*3)/7 .... 10 times-1*30/7 = -2/7 = 5 rem
Method II
Euler of 7 =6
when 10^6 is divided by 7 it will give 1 as a remainder
R= 10^10/7 =10^6 * 10^4/7 = 3^4/7
R = 4
Similarly all other terms will give remainder of 4 with 7
therefore total Remainder = 4+4+4+4....... (upto 10 terms)
which is 40 now 40 /7 = 5 which is the final remainder
Euler of 7 =6
when 10^6 is divided by 7 it will give 1 as a remainder
R= 10^10/7 =10^6 * 10^4/7 = 3^4/7
R = 4
Similarly all other terms will give remainder of 4 with 7
therefore total Remainder = 4+4+4+4....... (upto 10 terms)
which is 40 now 40 /7 = 5 which is the final remainder
P's present age is 4 yr. more than Q's age after six yrs. The sum of the present ages of P and Q together is 70 yr. What is the present age of Q ??
1.) 32 yr.
2.) 30 yr.
3.) 34 yr.
4.) 40 yr.
pls help me with the method to solve this sum..
yes cud u pls explain the method pls?
Just use the options..No need to solve using equations. i would personally take easiest of options first(which is round numbers). Ex if i took 30 then P's age is 40 (since sum is 70). now 1st line states P's present age 40 is 4 yrs more than q's age 6 yrs after which would be 36. Thus solved.
@Raghu5jan :
Let current age of P and Q be p,q respectively.
condition 1: p = (q+6)+4 = q+10 => p - q =10 .................(1)
condition 2: p + q = 70 .......................(2)
Solve 1 and 2 : 2p = 80 => p = 40 => q = 30.
two boats go downstream from point X to Y. the faster boat covers the distance from X to Y 1.5 times as fast as the slower boat. it is known that for every hour the slower boat lags behind the faster boat by 8 km. However if they go upstream then the faster boat covers the distance from Y to X in half time than slower boat. Find speed of faster boat in still water (please explain)???
a. 12 kmph b. 20 kmph c. 24 kmph d. 25 kmph