Find the least number which satisfies the following conditions:
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okay let me see if i can exlain this....its kinda hard
okay so we know that the sum of the digits is 1 more than a perfect square. so sum can only be 5, 10,17,26,37,50,65,82....from this 5 is too small and 37, 50, 65 and 82 are too large i.e. the number will be huge and problem will be almost impossible.
now keeping that in mind , lets consider its a 4 digit number
therefore the last 2 digits and the last digit should also be perfect squares (given in question). hence the last 2 digits can only be 81, 64, 25,36,49
also as mentioned in the question last 3 digits should also be perfect squares. but no 3 digit number that ends with 81 or 64 is a perfect square hence they can be rejected.
lets check 25 only 225 and 625 are three digit numbers that end with 25. therefore we can say that 225 and 625 may be the last three digits of the given number.let the last digit be 'x'
now as mentioned above their sum can only be 10,17,26 so
x+2+2+5 = 10 => x=1
therefore the number we get is 1225 which satisfies all the conditions