A-four-digit-number-has-the-following-properties-a-It-is-a-perfect-square-b-The-first-two-digits-are-equal-c-The-last-two-digits-are-equal-Find-all-such-numbers-

Question Number 13391 by Tinkutara last updated on 19/May/17

$$\mathrm{A}\:\mathrm{four}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{first}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{The}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{such}\:\mathrm{numbers}. \\ $$

Answered by RasheedSindhi last updated on 19/May/17

let the number is x+10x+100y+1000y =11x+1100y =11(x+100y) Since the number is perfect square it contains every prime factor twice. So 11×11=121 is the divisor of the required number. Let the number is 121m^2 Since the number is of four digits 121m^2 ≤9999 11m<99 Hence the square root of the number may be 11,22,33,...,99 On testing each of these numbers only 88 is such number whose square fulfills the requirements 88^2 =7744

$$\mathrm{let}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is} \\ $$$$\mathrm{x}+\mathrm{10x}+\mathrm{100y}+\mathrm{1000y} \\ $$$$=\mathrm{11x}+\mathrm{1100y} \\ $$$$=\mathrm{11}\left(\mathrm{x}+\mathrm{100y}\right) \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{perfect} \\ $$$$\mathrm{square}\:\mathrm{it}\:\mathrm{contains}\:\mathrm{every}\:\mathrm{prime} \\ $$$$\mathrm{factor}\:\mathrm{twice}.\:\mathrm{So} \\ $$$$\mathrm{11}×\mathrm{11}=\mathrm{121}\:\mathrm{is}\:\mathrm{the}\:\mathrm{divisor}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{required}\:\mathrm{number}. \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\:\mathrm{121m}^{\mathrm{2}} \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{of}\:\mathrm{four} \\ $$$$\mathrm{digits}\:\:\mathrm{121m}^{\mathrm{2}} \leqslant\mathrm{9999} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{11m}<\mathrm{99} \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{may}\:\mathrm{be} \\ $$$$\mathrm{11},\mathrm{22},\mathrm{33},…,\mathrm{99} \\ $$$$\mathrm{On}\:\mathrm{testing}\:\mathrm{each}\:\mathrm{of}\:\mathrm{these} \\ $$$$\mathrm{numbers}\:\mathrm{only}\:\mathrm{88}\:\mathrm{is}\:\mathrm{such}\:\mathrm{number} \\ $$$$\mathrm{whose}\:\mathrm{square}\:\mathrm{fulfills}\:\mathrm{the} \\ $$$$\mathrm{requirements} \\ $$$$\mathrm{88}^{\mathrm{2}} =\mathrm{7744} \\ $$

Commented by RasheedSindhi last updated on 19/May/17