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A-student-did-not-notice-that-the-multiplication-sign-between-two-7-digits-numbers-amd-wrote-one-14-digits-number-which-turned-out-to-be-3-times-the-would-be-product-What-are-the-initial-numbers-

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Question Number 142920 by PRITHWISH SEN 2 last updated on 07/Jun/21
A student did not notice that the multiplication sign between two 7−digits numbers amd wrote one 14−digits number which turned out to be 3 times the would be product. What are the initial numbers ?
$$\mathrm{A}\:\mathrm{student}\:\mathrm{did}\:\mathrm{not}\:\mathrm{notice}\:\mathrm{that}\:\mathrm{the}\:\mathrm{multiplication} \\ $$$$\mathrm{sign}\:\mathrm{between}\:\mathrm{two}\:\mathrm{7}−\mathrm{digits}\:\mathrm{numbers}\:\mathrm{amd}\:\mathrm{wrote} \\ $$$$\mathrm{one}\:\mathrm{14}−\mathrm{digits}\:\mathrm{number}\:\mathrm{which}\:\mathrm{turned}\:\mathrm{out}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{3}\:\mathrm{times}\:\mathrm{the}\:\mathrm{would}\:\mathrm{be}\:\mathrm{product}.\:\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{numbers}\:? \\ $$
Commented by PRITHWISH SEN 2 last updated on 07/Jun/21
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 07/Jun/21
only one solution: 1 666 667×3 333 334=((16 666 673 333 334)/3)
$${only}\:{one}\:{solution}: \\ $$$$\mathrm{1}\:\mathrm{666}\:\mathrm{667}×\mathrm{3}\:\mathrm{333}\:\mathrm{334}=\frac{\mathrm{16}\:\mathrm{666}\:\mathrm{673}\:\mathrm{333}\:\mathrm{334}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 07/Jun/21
say the 7 digit numbers are a and b. the 14 digit number is then ab_(−) =a×10^7 +b 3×a×b=a×10^7 +b b(3a−1)=a×10^7 3a−1=((10^7 )/k)=3a−1≥3×1000000−1 ⇒k≤3 3a−1=((10^7 )/k)=3a−1≤3×9999999−1 ⇒k≥1 with k=1: 3a−1=((10^7 )/1) ⇒a∉N with k=2: 3a−1=((10^7 )/2) ⇒a=1 666 667 ✓ ⇒b=ka=3 333 334 only solution is: 1666667×3333334=16666673333334/3
$${say}\:{the}\:\mathrm{7}\:{digit}\:{numbers}\:{are}\:{a}\:{and}\:{b}. \\ $$$${the}\:\mathrm{14}\:{digit}\:{number}\:{is}\:{then} \\ $$$$\underset{−} {{ab}}={a}×\mathrm{10}^{\mathrm{7}} +{b} \\ $$$$\mathrm{3}×{a}×{b}={a}×\mathrm{10}^{\mathrm{7}} +{b} \\ $$$${b}\left(\mathrm{3}{a}−\mathrm{1}\right)={a}×\mathrm{10}^{\mathrm{7}} \\ $$$$\mathrm{3}{a}−\mathrm{1}=\frac{\mathrm{10}^{\mathrm{7}} }{{k}}=\mathrm{3}{a}−\mathrm{1}\geqslant\mathrm{3}×\mathrm{1000000}−\mathrm{1}\:\Rightarrow{k}\leqslant\mathrm{3} \\ $$$$\mathrm{3}{a}−\mathrm{1}=\frac{\mathrm{10}^{\mathrm{7}} }{{k}}=\mathrm{3}{a}−\mathrm{1}\leqslant\mathrm{3}×\mathrm{9999999}−\mathrm{1}\:\Rightarrow{k}\geqslant\mathrm{1} \\ $$$${with}\:{k}=\mathrm{1}: \\ $$$$\mathrm{3}{a}−\mathrm{1}=\frac{\mathrm{10}^{\mathrm{7}} }{\mathrm{1}}\:\Rightarrow{a}\notin{N} \\ $$$${with}\:{k}=\mathrm{2}: \\ $$$$\mathrm{3}{a}−\mathrm{1}=\frac{\mathrm{10}^{\mathrm{7}} }{\mathrm{2}}\:\Rightarrow{a}=\mathrm{1}\:\mathrm{666}\:\mathrm{667}\:\checkmark \\ $$$$\Rightarrow{b}={ka}=\mathrm{3}\:\mathrm{333}\:\mathrm{334} \\ $$$$ \\ $$$${only}\:{solution}\:{is}: \\ $$$$\mathrm{1666667}×\mathrm{3333334}=\mathrm{16666673333334}/\mathrm{3} \\ $$

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