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Determine-the-number-of-five-digit-integers-37abc-in-base-10-such-that-each-of-the-numbers-37abc-37bca-and-37cab-is-divisible-by-37-




Question Number 19637 by Tinkutara last updated on 13/Aug/17
Determine the number of five-digit  integers (37abc) in base 10 such that  each of the numbers (37abc), (37bca)  and 37cab is divisible by 37.
$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{five}-\mathrm{digit} \\ $$$$\mathrm{integers}\:\left(\mathrm{37}{abc}\right)\:\mathrm{in}\:\mathrm{base}\:\mathrm{10}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}\:\left(\mathrm{37}{abc}\right),\:\left(\mathrm{37}{bca}\right) \\ $$$$\mathrm{and}\:\mathrm{37}{cab}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{37}. \\ $$
Answered by Rasheed.Sindhi last updated on 14/Aug/17
37abc=37000+abc  37∣37000+abc  37∣3700⇒37∣abc  ⇒abc is multiple of 37  As a may be 0 ,abc may be  2-digit or 3-digit multiple  of  37:  37×0,37×1,37×2,...,37×n  000,037,074,...,37n≤999 :37n  is last term GP.  n=[((999)/(37))]=27  000^(1) ,(037,074,...,37n^(n=27) ≤999)   1+27=28  Total multiples are 28 having  property given in the question.  For example  37=037⇒abc=037=37×1  bca=370=37×10, cab=703=37×19  So number 37037  Its variations: 37370 & 37703  37∣37037 , 3∣37370 , 37∣37703
$$\mathrm{37abc}=\mathrm{37000}+\mathrm{abc} \\ $$$$\mathrm{37}\mid\mathrm{37000}+\mathrm{abc} \\ $$$$\mathrm{37}\mid\mathrm{3700}\Rightarrow\mathrm{37}\mid\mathrm{abc} \\ $$$$\Rightarrow\mathrm{abc}\:\mathrm{is}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{37} \\ $$$$\mathrm{As}\:\mathrm{a}\:\mathrm{may}\:\mathrm{be}\:\mathrm{0}\:,\mathrm{abc}\:\mathrm{may}\:\mathrm{be} \\ $$$$\mathrm{2}-\mathrm{digit}\:\mathrm{or}\:\mathrm{3}-\mathrm{digit}\:\mathrm{multiple} \\ $$$$\mathrm{of}\:\:\mathrm{37}: \\ $$$$\mathrm{37}×\mathrm{0},\mathrm{37}×\mathrm{1},\mathrm{37}×\mathrm{2},…,\mathrm{37}×\mathrm{n} \\ $$$$\mathrm{000},\mathrm{037},\mathrm{074},…,\mathrm{37n}\leqslant\mathrm{999}\::\mathrm{37n} \\ $$$$\mathrm{is}\:\mathrm{last}\:\mathrm{term}\:\mathrm{GP}. \\ $$$$\mathrm{n}=\left[\frac{\mathrm{999}}{\mathrm{37}}\right]=\mathrm{27} \\ $$$$\overset{\mathrm{1}} {\mathrm{000}},\left(\overset{\mathrm{n}=\mathrm{27}} {\mathrm{037},\mathrm{074},…,\mathrm{37n}}\leqslant\mathrm{999}\right)\: \\ $$$$\mathrm{1}+\mathrm{27}=\mathrm{28} \\ $$$$\mathrm{Total}\:\mathrm{multiples}\:\mathrm{are}\:\mathrm{28}\:\mathrm{having} \\ $$$$\mathrm{property}\:\mathrm{given}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}. \\ $$$$\mathrm{For}\:\mathrm{example} \\ $$$$\mathrm{37}=\mathrm{037}\Rightarrow\mathrm{abc}=\mathrm{037}=\mathrm{37}×\mathrm{1} \\ $$$$\mathrm{bca}=\mathrm{370}=\mathrm{37}×\mathrm{10},\:\mathrm{cab}=\mathrm{703}=\mathrm{37}×\mathrm{19} \\ $$$$\mathrm{So}\:\mathrm{number}\:\mathrm{37037} \\ $$$$\mathrm{Its}\:\mathrm{variations}:\:\mathrm{37370}\:\&\:\mathrm{37703} \\ $$$$\mathrm{37}\mid\mathrm{37037}\:,\:\mathrm{3}\mid\mathrm{37370}\:,\:\mathrm{37}\mid\mathrm{37703} \\ $$
Commented by Tinkutara last updated on 14/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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