€Ž99(c - a) = 66k99(c - a) = 198, 396, 594, 792c - a = 2, 4, 6, 8for 2, 30 such numbersfor 4, 20 such numbersfor 6, 10 such numbersfor 8, 0 such numbersso, 60 such numbers
so u plz check ur ans again ..
the algebric expression of the number is 99(x-z). for this number to be divisible by 66, (x-z) has to be even. i.e only an even multiple of 99 will be divisible by 66. so (x-z) can be 2/4/6/8. It's given to us that 'z' is even.
1. (X-Z) TO BE 2 - Z can be 2, 4, 6 and x will be 4, 6, 8 respectively. Z cannot be 8 coz as x and z are digital numbers, if z becomes 8 then x has to be 10 which is not possible. HENCE THREE SUCH NUMBERS, NAMELY (4,2) (6,4) (8,6)
2. (X-Z) TO BE 4 - Z can be 2 and 4 and x will this be 6 and 4 respectively. Z cannot be 6 and 8 for the reason stated in point 1. hence two numbers namely (2,6) (4,8).
3. (X-Z) TO BE 6 - Z can only be 2 and x will be 8. more than that is not possible. hence just one number namely (2,8).
thus we get in total 6 such numbers. now for these 6 numbers, x can take any value ranging from 0-9 because as we saw eventually 10y and 10y gets cancelled and they do not contribute anything. hence total such numbers satisfying the above given condition are 6*10= 60 numbers. hope this helped you Priya.. :)
Let Z = {1, 11, 111, 1111, ..., 11111 up to 55 times} and T be a subset of Z such that the sum of no two elements of T is divisible by 3. The maximum possible number of elements in T is: a. 18 b. 19 c. 20 d. 37
I'm getting 37 as answer but the answer given is 20, please help!
@akansh_1 said:Let Z = {1, 11, 111, 1111, ..., 11111 up to 55 times} and T be a subset of Z such that the sum of no two elements of T is divisible by 3. The maximum possible number of elements in T is:a. 18 b. 19 c. 20 d. 37I'm getting 37 as answer but the answer given is 20, please help!
The elements are of 3 type:
3k -> 18 in total
3k + 1 -> 19 in total
3k + 2 -> 18 in total
For max possible no. of elements, we take all of type 3k+1, and one of type 3k.
@[521497:aadit4you] You can apply to all programs given in the CAT form. for some programs, there are different qualification criteria like Work Ex and Sector of work ex- just check online in the respective iim's website whether you are eligible for the course if you have any doubts.
for odd numbers, tn= 2^(n-1) + 1; so s(n=50) = summation[2^(n-1)] =50 =a; it is a gp so its sum is: 1 (4^50 - 1)/4-1 + 50 similarly the sum for even n is 2 (4^50 -1)/4-1 - 50; . adding both of them wee get sum = 4^50 -1; and to obtain its remainder, we will obtain remainder of 4^50, and will subtract 1 from that remainder; so remainder of 2^100/127 is 16 and subtracting 1 we get 15.. but it is not in options..
The series of no.s (1, 1/2, 1/3, 1/4......1/1972) is taken. Now two no.s are taken from this series(the first two) say x, y. Then the operation x+y+x.y is performed to get a consolidated no. The process is repeated. What will be the value of the set after all no.s are consolidated into one number.
@surya the bacon said:The series of no.s (1, 1/2, 1/3, 1/4......1/1972) is taken. Now two no.s are taken from this series(the first two) say x, y. Then the operation x+y+x.y is performed to get a consolidated no. The process is repeated. What will be the value of the set after all no.s are consolidated into one number.
@surya the bacon said: The series of no.s (1, 1/2, 1/3, 1/4......1/1972) is taken. Now two no.s are taken from this series(the first two) say x, y. Then the operation x+y+x.y is performed to get a consolidated no. The process is repeated. What will be the value of the set after all no.s are consolidated into one number.
for odd numbers, tn= 2^(n-1) + 1; so s(n=50) = summation[2^(n-1)] =50 =a; it is a gp so its sum is: 1 (4^50 - 1)/4-1 + 50 similarly the sum for even n is 2 (4^50 -1)/4-1 - 50; . adding both of them wee get sum = 4^50 -1; and to obtain its remainder, we will obtain remainder of 4^50, and will subtract 1 from that remainder; so remainder of 2^100/127 is 16 and subtracting 1 we get 15.. but it is not in options..
could not understand @ @[599204:sandip.yadav] plz xplain summation[2^(n-1)] =50 =a;???????????????
