If $$\begin{align}2*6&=7,\\4*2&=72\quad\text{and}\\7*3&=144\end{align}$$
then,
$$\;\;5*8=\;?\qquad\qquad$$
$\begingroup$
$\endgroup$
6
-
$\begingroup$ Is it relevant that 144 vf n ahzore va gur Svobanppv frdhrapr. $\endgroup$– jsmCommented May 3, 2018 at 15:35
-
3$\begingroup$ the others aren't so why would it matter if 144 is? Just curious about your thinking... $\endgroup$– Dr tCommented May 3, 2018 at 16:09
-
$\begingroup$ what exactly...did @Mike Earnest just edit here that lends clarity and meaning to the initial post? I didn't see any change. Please explain. $\endgroup$– Dr tCommented May 3, 2018 at 16:33
-
$\begingroup$ @Drt I removed the [logical-deduction] tag and replaced it with [pattern], since this puzzle cannot be solved by logical inference alone. $\endgroup$– Mike EarnestCommented May 3, 2018 at 16:37
-
6$\begingroup$ Your multiplication is wrong. $\endgroup$– RajanCommented May 5, 2018 at 11:58
|
Show 1 more comment
3 Answers
$\begingroup$
$\endgroup$
Not sure about any mathematic operation but can find this pattern(could be wrong!):
$2∗6 = 7$;
$4∗2 = 72$; 7 from RHS of 1st row and 2 1st number in 1st row.
$7∗3 = 144$; 7*2 from RHS of 2nd row(multiply) and 4 1st number in 2nd row.
So,
$5∗8 = 167$; $1*4*4$ from RHS of 3rd row and 7 1st number in 3rd row.
edit: just noticed i mistaken 7 with 17, so corrected;)
$\begingroup$
$\endgroup$
I found the same pattern as @Preet, but had a different result:
$2 * 6 = 7 \Rightarrow 7 $; initial statement
$4 * 2 = 72 \Rightarrow 7$ ~ $2 $; product of previous result (just 7) & first row (2)
$7 * 3 = 144 \Rightarrow 14$ ~ $4 $; product of previous result $(7*2 = 14)$ and first row (4)
$5 * 8 = 567 \Rightarrow 56$ ~ $7 $; product of previous result $(14*4 = 56)$ and first row (7)
So the answer might be
$567.$
$\begingroup$
$\endgroup$
4
Strip out the symbols and multiply by 16 to find the next row
2*6=7 => 267 * 16 = 4272
4*2=72 => 4272 * 16 = 73144
7*3=144 => 73144 * 16 = 581704
therefore: 5*8=1704
-
4$\begingroup$ $4272\cdot16=68352$ and $73144\cdot16=1170304$? Or just note that the products of the units digits don't even match? Or that $581703$ is odd? $\endgroup$– noedneCommented May 4, 2018 at 19:35
-
-
4$\begingroup$ How are you obtaining these numbers? $\endgroup$– noedneCommented May 4, 2018 at 19:56
-
6$\begingroup$ Yes, I agree with @noedne, the second and third products aren't correct. $\endgroup$– hexominoCommented May 4, 2018 at 20:06