Doubly vertically filled in

Logic Level 3

□ 8 × □ 7 1 □ 6 □ 2 □ □ 3 □ □ \large{\begin{array}{ccccccc} && & & & &\square &8 \\ \times && & & & & \square &7 \\ \hline && & & & 1 & \square &6 \\ && & & \square & 2 & \square & \\ \hline && & & \square & 3 &\square & \square \\ \hline \end{array}}

The above represents an incomplete long multiplication. What is the sum of all the missing digits?

Clarification: The above shows the product of two 2-digit integers that yields a 4-digit integer.


The answer is 32.

Pi Han Goh by Pi Han Goh , 30, Malaysia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Brock Brown
Dec 15, 2016

First, consider that because we're using a long multiplication method: ? 8 × 7 = 1 ? 6 \boxed{?}8\times7=1\boxed{?}6

18 is the biggest number we can multiply 7 by before exceeding the largest possible product 1 9 6 1\boxed{9}6 , therefore

1 8 × 7 = 1 2 6 \boxed{1}8\times7=1\boxed{2}6

From there we know that

18 × ? 7 = ? 3 ? ? 18\times\boxed{?}7=\boxed{?}3\boxed{?}\boxed{?}

If you iterate you can see that 18 × 7 7 18\times\boxed{7}7 is the only four digit number where the 2nd digit is 3.

18 × 7 7 = 1 3 8 6 18\times\boxed{7}7=\boxed{1}3\boxed{8}\boxed{6}

Finally, we can see that 126 + ? 2 ? 0 = 1386 126+\boxed{?}2\boxed{?}0=1386 ⟹ 126 + 1 2 ? 0 = 1386 \implies 126+\boxed{1}2\boxed{?}0=1386 ⟹ 126 + 1 2 6 0 = 1386 \implies 126+\boxed{1}2\boxed{6}0=1386

Add all of your newfound missing answers, and we arrive at 32 \boxed{32}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Oh, I'm slightly disappointed that I didn't see a Python solution =P

Pi Han Goh - 4 years, 6 months ago

Log in to reply

Well, you know, funny you should say that... it turns out that when you try every combination of numbers it gives you 5 different possible values for the expression: {1316, 1326, 1386, 2376, 3306}

I tried brute force Python, Pi, I honestly did. T.T 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
 
from itertools import product
digits = list(str(i) for i in range(10))
possible = set()
for a, b in product(digits, repeat=2):
    try:
        test = eval("{0}8*{1}7".format(a,b))
        possible.add(test)
    except SyntaxError: pass
new_possible = set()
for a,b,c in product(digits, repeat=3):
    try:
        test = eval("1{0}6+{1}2{2}0".format(a,b,c))
        if test in possible:
            new_possible.add(test)
    except SyntaxError: pass
possible = new_possible
new_possible = set()
for a,b,c in product(digits, repeat=3):
    try:
        test = eval("{0}3{1}{2}".format(a,b,c))
        if test in possible:
            new_possible.add(test)
    except SyntaxError: pass
possible = list(new_possible)
possible.sort()
print (possible)

Brock Brown - 4 years, 6 months ago

Log in to reply

I couldn't have asked for a better answer. This is well articulated.

Pi Han Goh - 4 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...