Here is a table, inside there are simple operations to perform, but there are also particular numbers that don't have the same logic.
So how difficult will it be to find the last number and the pattern to solve this one?
Hope you'll enjoy this!
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$\begingroup$ The accepted answer fits a pattern established by only 4 of the 35 given numbers and not even all the numbers of the diagonal? An answer of "100" would at least fit a pattern established by 30 of the given numbers. This really looks like each of the white cells is meant to take into account all of the cells to the left and above it, including shaded cells. $\endgroup$– humnMay 1 '20 at 2:53
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105. Each diagonal cell is the multiplication entry $+25, +20, +15, +10, +5$.
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$\begingroup$ Yeah, that's what I thought: the diagonal line is composed of every primary value, multiplied with itself (that means number raised at 2), plus 5 multiplied for the quantity of rows remaining under it plus 1 The result explained with a formula is: Result = [ n^2 + 5 * (rows_under + 1) ] $\endgroup$– xKobaltApr 30 '20 at 15:58
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Answer could be 95 or 113 or 112. Here's how: a) 95: After 29, all the numbers in that diagonal are moving like, 1st digit is next odd number and 2nd digit is obtained by subtracting the first digit number from the 2nd digit of predecessor diagonal number. So, numbers goes like, 2 9 3 6 i.e.(9-3) 5 1 i.e.(6-5) 7 4 i.e.([1]1 - 7) 9 5 i.e. ([1]4 - 9)
b) 113 or 112: Same logic as above, with a difference that 1st numbers are successive prime numbers and not odd, which will make the first number 11. For second number, there could be 02 options either treat 1st digit as 11 [eleven] or 11[one and one]. If we treat it, as 11 [eleven], no. is 11 3 i.e. ([1]4 - 11) 11 [one and one], no. is 11 2 i.e. ([1]4 - (1 + 1)) Among 113 or 112, 113 sounds more logical, cause we have taken first digit as 11 [eleven].
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$\begingroup$ Very tricky... it's not the solution I thought, but I have appreciated it $\endgroup$– xKobaltApr 30 '20 at 15:49
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The question is ''105'' because ''29+7=36'' ''36+7+7+1=51'' ''51+7+7+7+2=74'' and ''74+7+7+7+7+3= 105''
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$\begingroup$ I didn't think about that before, it seems tricky, but after I've done some reasoning it works $\endgroup$– xKobaltApr 30 '20 at 13:41
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$\begingroup$ Nice - this is equivalent. You've basically got the recurrence relation: $x_k = x_{k-1} + 7(k-1) + (k-2)$ with the initial value of $x_1 = 29$, and where $k=n/2$ for the $n$ in JMP's formula in their comment on the accepted answer (so $k$ is the column counting 1, 2, 3 etc, whereas $n$ is the value in the pale blue row, counting 2, 4, 6 etc). This recurrence relation can be solved as $x_k = 4k^2 - 5k +30$ which is equivalent to JMP's formula above once you've made the substitution for $n$. $\endgroup$– TimMay 1 '20 at 11:09