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NVAMediumJEE 2023Complex Numbers Basics

JEE Mathematics 2023 Question with Solution

5050th root of a number xx is 1212 and 5050th root of another number yy is 1818. Then the remainder obtained on dividing x+yx + y by 2525 is .....

Answer

Correct answer:23

Step-by-step solution

Standard Method

Given: x=1250x = 12^{50} and y=1850y = 18^{50}.

Find: The remainder when x+yx + y is divided by 2525.

We need to compute x+y(mod25)x+y \pmod{25}.

First,

122=144≡19(mod25)12^2 = 144 \equiv 19 \pmod{25} 124=(122)2≡192=361≡11(mod25)12^4 = (12^2)^2 \equiv 19^2 = 361 \equiv 11 \pmod{25} 128=(124)2≡112=121≡21(mod25)12^8 = (12^4)^2 \equiv 11^2 = 121 \equiv 21 \pmod{25} 1216=(128)2≡212=441≡16(mod25)12^{16} = (12^8)^2 \equiv 21^2 = 441 \equiv 16 \pmod{25} 1232=(1216)2≡162=256≡6(mod25)12^{32} = (12^{16})^2 \equiv 16^2 = 256 \equiv 6 \pmod{25}

So,

1250=1232⋅1216⋅122≡6⋅16⋅19(mod25)12^{50} = 12^{32} \cdot 12^{16} \cdot 12^2 \equiv 6 \cdot 16 \cdot 19 \pmod{25} 6⋅16=96≡21(mod25)6 \cdot 16 = 96 \equiv 21 \pmod{25} 21⋅19=399≡24(mod25)21 \cdot 19 = 399 \equiv 24 \pmod{25}

Hence,

1250≡24(mod25)12^{50} \equiv 24 \pmod{25}

Now,

182=324≡24(mod25)18^2 = 324 \equiv 24 \pmod{25} 184=(182)2≡242=576≡1(mod25)18^4 = (18^2)^2 \equiv 24^2 = 576 \equiv 1 \pmod{25}

Therefore higher powers repeat accordingly, and

1850=1832⋅1816⋅182≡1⋅1⋅24(mod25)18^{50} = 18^{32} \cdot 18^{16} \cdot 18^2 \equiv 1 \cdot 1 \cdot 24 \pmod{25}

So,

1850≡24(mod25)18^{50} \equiv 24 \pmod{25}

Therefore,

x+y≡24+24=48≡23(mod25)x+y \equiv 24+24 = 48 \equiv 23 \pmod{25}

So, the remainder obtained is 2323.

Using binomial form

Given: x=1250x = 12^{50} and y=1850y = 18^{50}.

Find: The remainder when x+yx+y is divided by 2525.

Write

12=25−1312 = 25 - 13

and

18=25−718 = 25 - 7

A quicker observation from the extracted solution is that powers modulo 2525 settle to values near −1-1:

182=324≡−1(mod25)18^2 = 324 \equiv -1 \pmod{25}

Hence,

1850=(182)25≡(−1)25=−1≡24(mod25)18^{50} = (18^2)^{25} \equiv (-1)^{25} = -1 \equiv 24 \pmod{25}

Also, from repeated squaring,

1250≡24(mod25)12^{50} \equiv 24 \pmod{25}

Therefore,

x+y≡24+24=48≡23(mod25)x+y \equiv 24+24 = 48 \equiv 23 \pmod{25}

Thus, the remainder is 2323.

Common mistakes

  • Treating “5050th root of xx is 1212” as x=5012x = 50^{12}. This is incorrect because the statement means x=1250x = 12^{50}. Always convert root statements into exponential form carefully.

  • Finding 125012^{50} and 185018^{50} directly before dividing by 2525. This is unnecessary and impractical. Use modular arithmetic at each step to keep the numbers small.

  • Forgetting to reduce the final sum modulo 2525. Even after getting 1250≡2412^{50} \equiv 24 and 1850≡2418^{50} \equiv 24, the sum is 4848, and the required remainder is 48 mod 25=2348 \bmod 25 = 23, not 4848.

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