Sauarv's Square (Level-1)

Level pending

Let ABC be a 3 digit number such that (ABC)^2 = XYZABC. Find the number of solutions of ABC ?

Details and Assumptions : ABC is a 3 digit number and not multiplication of A, B & C.
XYZABC is a 6 digit number where X,Y,Z,A,B,C may be equal or different.


The answer is 2.

Saurav Pal by Saurav Pal , 26, India

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1 solution

Saurav Pal
Mar 26, 2014

There are only 2 solutions of ABC : 376 & 625.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Aur Saurav, kaisa hai! Peheli baat toh tu LaTeX use kara seekh, aur doosri baat yeh ki agar tujhe kisi savaal ko solve karne ka tareeka nati aata, toh usko post mat kiya kar!!

Satvik Golechha - 7 years, 2 months ago

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Mujhe solve karne me aalas aata hai isliye solve nahi karta. There is a trick to solve to solve this question.

Saurav Pal - 7 years, 2 months ago

Answer to mera bhi sahi hai par solve karne ka tarika to bata

Shubhendra Singh - 7 years, 2 months ago

Can you explain how you concluded that there are only 2 solutions?

Calvin Lin Staff - 7 years, 2 months ago

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Firstly, 376+625=1001=10^3+1
If you see the unit digit of 376 and 625, and if you square that, the last digit of the number will be 6 and 5 respectively.
similarly, If you see the last 2 digits of the numbers and square them, you will get
76^2=5776. Last 2 digits are also 76.
25^2=625. last 2 digits are also 25.
also, 5+6=11
25+76=101
625+376=1001 There will be either 1 or 2 numbers and will sum up to 10^N+1.where N is the number of digits in that number.


Saurav Pal - 7 years, 2 months ago

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Good solution!

Parag Zode - 6 years, 5 months ago

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