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A six digit number begins with one. But if the one put at the other end, the resulting six digit number is 3 times greater than the original number. Find the number.

Aug 23, 2019

#1
+25275
+3

A six digit number begins with one.

But if the one put at the other end, the resulting six digit number is 3 times greater than the original number.

Find the number.

$$\text{The six digit number begins with one: \mathbf{1rstuv} } \\ \text{The resulting six digit number ends with one: \mathbf{rstuv1} }$$

$$\begin{array}{|lcll|} \hline \mathbf{rstuv1} = 3\cdot \mathbf{1rstuv} \\ \hline \end{array}$$

$$\small{ \begin{array}{|rcll|} \hline r\cdot 10^5+s\cdot 10^4+t\cdot 10^3+u\cdot 10^2+v\cdot 10+1 &=& 3\cdot (1\cdot 10^5+r\cdot 10^4+s\cdot 10^3+t\cdot 10^2+u\cdot 10+v ) \\\\ r\cdot 10^5+s\cdot 10^4+t\cdot 10^3+u\cdot 10^2+v\cdot 10+1 &=& 3\cdot 10^5+ r\cdot3\cdot 10^4+ s\cdot 3\cdot 10^3+t\cdot 3\cdot 10^2+ u\cdot 3\cdot 10+3\cdot v \\\\ r(\cdot 10^5-\cdot3\cdot 10^4) +s(\cdot 10^4-\cdot 3\cdot 10^3) +t(\cdot 10^3-\cdot 3\cdot 10^2 ) +u(\cdot 10^2-\cdot 3\cdot 10 ) +v(\cdot 10+1-3) &=& 3\cdot 10^5-1 \\\\ 70000r+7000s+700t+70u+7v &=& 299999 \\ 7\cdot r\cdot 10^4 +7\cdot s\cdot 10^3 +7\cdot t\cdot 10^2 +7\cdot u\cdot 10 +7v &=& 299999 \\ 7\cdot \left(r\cdot 10^4+s\cdot 10^3 +t\cdot 10^2 +u\cdot 10 +v \right) &=& 299999 \quad | \quad : 7 \\ r\cdot 10^4+s\cdot 10^3 +t\cdot 10^2 +u\cdot 10 +v &=& 42857 \\ \mathbf{ r s t u v } &=& \mathbf{ 42857 } \\ \hline \end{array} }$$

The number is $$\mathbf{142857}$$

Check: $$3\cdot 142857 = 428571$$

Aug 23, 2019
edited by heureka  Aug 23, 2019
#2
+111396
+1

Thanks, heureka.....that's a neat solution   !!!

CPhill  Aug 23, 2019
#3
+25275
+2

Thank you, CPhill !

heureka  Aug 23, 2019