R * RR

Logic Level 1

R × R R P Q R \begin{array} { l l l } & & & R \\ \times & & R & R \\ \hline & P & Q& R \\ \end{array}

How many solutions are there to the above cryptogram ?

Note: The numbers are not allowed to start with 0.

0 1 2 4

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2 solutions

Angela Fajardo
Nov 2, 2016

If you will notice when R × R happens, the result is also R \large \text{If you will notice when } R\times R \text{ happens, the result is also R}

In order to lessen the possible numbers, we are only concerned with the ones digit of the number \large \text{In order to lessen the possible numbers, we are only concerned with the ones digit of the number}

1 × 1 = 1 2 × 2 = 4 3 × 3 = 9 4 × 4 = 1 6 5 × 5 = 2 5 6 × 6 = 3 6 7 × 7 = 4 9 8 × 8 = 6 4 9 × 9 = 8 1 \large \begin{matrix} 1 & \times & 1 & = & \boxed { 1 } \\ 2 & \times & 2 & = & \boxed { 4 } \\ 3 & \times & 3 & = & \boxed { 9 } \\ 4 & \times & 4 & = & 1\boxed { 6 } \\ 5 & \times & 5 & = & 2\boxed { 5 } \\ 6 & \times & 6 & = & 3\boxed { 6 } \\ 7 & \times & 7 & = & 4\boxed { 9 } \\ 8 & \times & 8 & = & 6\boxed { 4 } \\ 9 & \times & 9 & = & 8\boxed { 1 } \end{matrix}

So, given the numbers 1 to 9 only 1, 5, and 6 fits the criteria \large \text{So, given the numbers 1 to 9 only 1, 5, and 6 fits the criteria}

R R × R = X Y Z 11 × 1 = 11 55 × 5 = 275 66 × 6 = 396 \large \begin{matrix} RR & \times & R & = & XYZ \\ 11 & \times & 1 & = & 11 \\ 55 & \times & 5 & = & 275 \\ 66 & \times & 6 & = & 396 \end{matrix}

And from these, we can see that only R = 5 and R = 6 satisfies the cryptogram. \large \text{And from these, we can see that only R = 5 and R = 6 satisfies the cryptogram.}

Observing the last digit gives us the information :)

Chung Kevin - 4 years, 7 months ago

Ah yes, sometimes we can just run through all possibilities for a digit, and see what the outcomes are :)

Calvin Lin Staff - 4 years, 7 months ago
Mohammad Khaza
Jul 7, 2017

two solutions are: 5x 55= 275

and 6 x 66 =396

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