In the Memory of Maryam Mirzakhani

$\large{\begin{array}{ccccccc} &&&&n_5 & n_4& n_3&n_2 & n_1& \\ &&&& & Q& U&E & E&N\\ &&&& & & & & O&7\\ &&&& & Q& U&E & E&N\\ &&&& & & & & O&7\\ +&&&1&3 & 7& E& N & 3&E\\ \hline &&& {\color{cyan}M}&{\color{cyan}A} & {\color{cyan}4}& {\color{cyan}0}& {\color{cyan}A} & {\color{cyan}M}&M \end{array}}$

There is no carry over in the millions place. Thus $\color{cyan}M=1$ and in the ones and tens places we must get an odd sum. Thus $n_1=2, 4$ .

If $n_1=4$ , then $2N+14+E=41$ and $2N+E=27$ . It follows that $N=E=9$ . Tens place gives a sum as $25+2O$ , which can be $31$ or $41$ . So $n_2=3$ or $4$ . Then hundreds place's sum will be $n_2+2E+N=n_2+27=30$ or $31$ . Hence $A=0$ or $1$ . But this is a contradiction since $A>3$ .

So we must have $n_1=2$ . Thus $2N+E=7$ . Remember that $E$ must be odd.

In the thousands place, $n_3$ must be an odd number to give an even sum and since $n_2+2E+N<30$ , then $n_3=1$ .

Now, lets go to to the ten thousands place, where $n_4$ must be an odd number. Notice that the sum of thousands place is $n_3+2U+E<30$ . Therefore, $n_4=1$ , and for the ten thousands place, we get $2Q+8=14$ or $24$ .

• If $2Q+8=14$ , then $\color{cyan}Q=3$ and $\color{cyan}A=4$ . Hence sum of hundreds place is $n_2+2E+N=14$ and $n_2+2(7-2N)+N=14$ . It follows that $n_2=3N$ . Thus, we must have $n_2=3$ and $\color{cyan}N=1$ . Then $\color{cyan}E=5$ . From tens place $15+2O=31$ , which gives $\color{cyan}O=8$ . Finally from thousands place, we get $2U+E=9$ and $\color{cyan}U=2$ .

• If $2Q+8=24$ , then $Q=8$ and $A=5$ . Hence sum of hundreds place is $n_2+2E+N=15$ and $n_2+2(7-2N)+N=15$ . It follows that $n_2=3N+1$ . Thus, we must have $n_2=4$ and $N=1$ . Then $E=5$ . So from tens place, we get $15+2O=41$ , and $O=13$ , which is impossible.

Thus, the answer is $Q+U+E+N+O+F+S+C+I+M+A+R+Y=3+2+5+1+8+7+1+3+7+1+4+4+0=\boxed{46}$ .

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If you remove each one of the constraints, problem becomes extremely difficult! I think more beautiful variations of this cryptogram can be made, with changing constraints!

at first i took the values-----RY=40; CI=37; SF=17

then i did permutation with other numbers one by one.

the values could be:

Q=3 ; U=2; E=5; N=1 ; O=8 ; F=7 ; S=1[as SF=17, FS should be=71]; C=3 ; I=7; M=1; A=4; R=4; Y=0[given that RY=40]

so, now the summation is= $3+2+5+1+8+7+1+3+7+1+4+4+0$

= $46$