Quote:
Originally Posted by HPK
1) We can reduce it to 7^4 .. So it will be 49^2 ie. 2401..
Ans : 0
2) Ans : 29 ..m doubtful bout this 1..pls clarify..
3) Ans : 7
Coz.. any no. ending in 1 and 5 will hv their higher powers ending in 1 n 5 respectively
if u consider the cube of 3...its 27...so the no will end with 7...
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@HPK,
Here are the answers
37^4 = (30+7)^4...now we use binomial theorem to find the tens place...since all the number till 4c2*30^2*7^2 will have two zeros in the end we'll consider only 4c3*30*7^3 + 7^4 to find the tens digit...
to make the calculation easier we'll consider only th last two digits in all calc...
4c3*30*7^3 = 4*30*343 = 120*343 = 12*343*10= ends with 60 (since 3*2=6)
7^4 = 343*7 = ends with 01
therefore 60+01=61....therefore tens digit is 6...
Which is the smallest two dogit number which will never be a perfect square in any base?
1) 29 2) 32 3) 41 4)62
The best way would be to convert the given number to base 10 and check if it is a perfect squuare or not. But that would involve a lot of trial and error stuff, but still I think it would not take a lot of time.
In this process of base conversion, the units digit of the number remains intact in the bases under consideration( ofcourse except the cases in which a digit doesn’t exist in a particular base e.g 9 doesn’t exist in base
. So if subtract the unit digit from a perfect sqaure then the remainder should be an integral multiple of the tens digit of the number in the base under consideration.
e.g. Consider option a) 29
Lets consider a perfect square say 225. When we subtract 9 from it, the remainder that we get is 116. Now if this remainder is an integral multiple of the tens digit (which in this case is 2), then the number is a perfect square in that base (i.e 10
225 =
2*108 +
9.
Similarly find out if others can fit in such expansions or not.
25 =
4*6 +
1. (option c can be a square)
If we consider option b) 32, if we subtract
2 from any square say 64, then the remainder is not an integral multiple of
3. This would hold true for any square(not just 64). Thus 32 cannot be a square in any base. And obviously, 62 cannot be a square either.
3. Answer is 7.
4 2) P(x) is a polynomial defined for all real values of x. P2 (x) = P(P(x)), P3 (x) = P(P2 (x)) so on and so forth.The equation P5 (x) + P(x) = 34x + 44. Then find the value of P(37) - P(5).
a) 32 b) 64 c) 128 d) 256
SOL:-
P5 (x) + P(x) = 34x + 44
let P(x)=ax+b ... 'ax+b' since {P5 (x) + P(x)} is in the same format then P2(x) = P(P(x)) = a(ax+b)+b =a^2x+(ab+b) we observe that for P2(x) power of x is 2...follow on similar lines P5(x) will have power of 5 i.e. a^5 rest of the terms are constant...therefore in P5 (x) n P(x) the terms having x will be a^5 n a respectively therefore a^5x+ax=34x => a^5+a=34 => a^5+a=32+2 => a=2
therefore P(x)=2x+b...now start solving
P2(x) = P(P(x)) = 2(2x+b)+b = 4x+3b
P3(x) = P(P2(x)) = 2(4x+3b)+b = 8x+7b
P4(x) = P(P4(x)) = 2(8x+7b)+b = 16x+15b
P5(x) = P(P4(x)) = 2(16x+15b)+b = 32x+31b
therefore P5 (x) + P(x)= {32x+31b} + 2x+b = 34x + 44 => 32b=44 => b=11/8
therefore P(x)= 2x+(11/
...now sub the values of x given in the question
P(37)-P(5)= {2(37)+(11/
} - {2(5)+(11/
} = 74-10 = 64...ans
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