@[510729:erm]: i got 72 as the answer. i jus solved the long way since i din know the exact trick.
@[595801:geetlove]: Just keep multiplyin the last digit and u get like 76^2 which has the last 2 digits as 76.
@[595801:geetlove]: i think the last term would remain like
1/4(1/3-1/7+1/7-1/11+1/11-1/15....) so unless am wrong u shud know the last term to find the answer.
it should be 1/4*(1/3-1/x).
correct me if i am wrong :)
1 what is unit of 7^11^22^33??
2 what is the sum of digits of the least multiple of 13 , which when diveded by 6, 8 & 12 leaves a remainder of 5,7,11 as the remainder????
PLZ elaborate your method of solving
1 and 8 are the first two natural no.s having 1+2+3+...n= perfect square, which is the fourth such number?...xplain the solution as well..
@erm said:1 what is unit of 7^11^22^33?? 2 what is the sum of digits of the least multiple of 13 , which when diveded by 6, 8 & 12 leaves a remainder of 5,7,11 as the remainder???? PLZ elaborate your method of solving
Ans to 1st is 7
We have to check the remainder when 11^22^33 is divided by 4
Since its 1 hence unit digit will be same as 7^1
For 2nd No will be of the form LCM(6,8,12)*n-1
But we have to take the multiple of 13
Hence lowest no will be 143 for n=6
Sum of digits = 8
We have to check the remainder when 11^22^33 is divided by 4
Since its 1 hence unit digit will be same as 7^1
For 2nd No will be of the form LCM(6,8,12)*n-1
But we have to take the multiple of 13
Hence lowest no will be 143 for n=6
Sum of digits = 8
hi .... need to know if growth has been asked in a question .... then does it mean the magnitude between the initial and the final value or it simply the percentage increase
Actually I got confused because of the answer options ... In the answer options they have written Which company has the higher growth than the previous month ?Based on this question i calculated the percentage growth . But the answer was given on the basis of magnitude of the growth ...
@priya.patil said:@sandip.yadav bt OA is 60 can u plzz verify ur ans again..
the answer is 100,, as y can also vary from 0 - 9,, so the resulting answer is 10 * 10 = 100
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how can it be 60..??
if u have proof then show it.
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how can it be 60..??
if u have proof then show it.
@[599204:sandip.yadav] are i dont kw d OA n sir had told its 60 woh toh i also dint understood how he got that anyways i dont hv d proof n i was jst asking for help use mein proof kaha se aaya n thanx i will check it again
An intelligence agency forms a two-digit code consisting of distinct digits selected from 0 through 9 such that the first digit is not zero. Some codes, when handwritten on a slip, can however potentially create confusion when read upside down - for example 61 may appear as 19. Find the number of codes for which there is no confusion.
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28
On their birthday, Ram and Shyam, a pair of twins, were gifted a large packet of chocolates, containing chocolates of four different flavours. They started eating the chocolates alternately, one chocolate at a time. What should be the minimum total number of chocolates that both of them together should have eaten before you can say that at least one of them has certainly eaten at least four chocolates of a single flavour?
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@harddev said:1 and 8 are the first two natural no.s having 1+2+3+...n= perfect square, which is the fourth such number?...xplain the solution as well..
the third term of this series is 49, and the fourth term of this series is 288..
proof:
we have sn= n(n+1)/2, and this has to be a square, then sn = a^2.
nw if n is an even integer then n= 2t, makin sn = t(2t +1)
or if n is odd then n +1 is even and then let n+1 = 2t, makin sn = (2t-1)t
so a^2 = t ( 2t + - 1 ); nw the two factors of a square are coprime to each other and thus share no factors between them,, so to be lhs a perfect square both factors shud be perfect squares by themselves... so by taking only the square values of t and generating the other factor, we can guess the further terms of this series..
proof:
we have sn= n(n+1)/2, and this has to be a square, then sn = a^2.
nw if n is an even integer then n= 2t, makin sn = t(2t +1)
or if n is odd then n +1 is even and then let n+1 = 2t, makin sn = (2t-1)t
so a^2 = t ( 2t + - 1 ); nw the two factors of a square are coprime to each other and thus share no factors between them,, so to be lhs a perfect square both factors shud be perfect squares by themselves... so by taking only the square values of t and generating the other factor, we can guess the further terms of this series..
@pathetic: the answer is 70.
@[599204:sandip.yadav] here goes my proof
€Ž99(c - a) = 66k
99(c - a) = 198, 396, 594, 792
c - a = 2, 4, 6, 8
for 2, 30 such numbers
for 4, 20 such numbers
for 6, 10 such numbers
for 8, 0 such numbers
so, 60 such numbers so u plz check ur ans again ..:)
@[244788:adirash20] thanx hv gone thro these in past old days on you tube... those videos r awesome
@erm said:1 what is unit of 7^11^22^33?? 2 what is the sum of digits of the least multiple of 13 , which when diveded by 6, 8 & 12 leaves a remainder of 5,7,11 as the remainder???? PLZ elaborate your method of solving
for ur first query, the answer is 07..
as 7raise to powers move in cycle of- 07, 49, 43, 01, 07, 49... in cycle of 4..
so if we have last two digits of its power we can guess its last two digits,
now, digits of 11 raise to something can be guessed by watching only the unit digit of power,, suoppose it is 3, then last two digits are 31, if 4 then 41.. etc..
now we have to find out the unit digit of 22^33..
so it slast digit is 2, which could be written as, 2^(32+1) = 4^16 * 2;
now unit digit of 4 is 6 if power is even, and 4 when power is odd.. so required unit digit is.. 6 * 2 = 2,
and 11^2 = 21,
and 7^21 = 07
as 7raise to powers move in cycle of- 07, 49, 43, 01, 07, 49... in cycle of 4..
so if we have last two digits of its power we can guess its last two digits,
now, digits of 11 raise to something can be guessed by watching only the unit digit of power,, suoppose it is 3, then last two digits are 31, if 4 then 41.. etc..
now we have to find out the unit digit of 22^33..
so it slast digit is 2, which could be written as, 2^(32+1) = 4^16 * 2;
now unit digit of 4 is 6 if power is even, and 4 when power is odd.. so required unit digit is.. 6 * 2 = 2,
and 11^2 = 21,
and 7^21 = 07
@priya: and for a = 0; 40 such numbers are possible, so 60 + 40 = 100;
as 200 is also an even three digit number which is subtracted by its reverse. gives 198, a number divisible by 66... 😛
Q. What is the remainder when 2^90 is divided by 91?
Ans given is 64 but I'm getting 1 as the answer. Puys please help.
....
According to Fermat's Little theorem,
a^p/p yields a remainder of a , if a and p are co-prime.
So, a^(p-1)/p will give 1 as remainder.
In this case, 2^(91-1)/91 will give 1 as remainder. Where I'm going wrong?
@akansh_1 said:Q. What is the remainder when 2^90 is divided by 91?Ans given is 64 but I'm getting 1 as the answer. Puys please help.....According to Fermat's Little theorem,a^p/p yields a remainder of a , if a and p are co-prime.So, a^(p-1)/p will give 1 as remainder.In this case, 2^(91-1)/91 will give 1 as remainder. Where I'm going wrong?