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enter image description here Each white square should be filled with a digit or one of these operators: +, -, ×, ÷. The numbers in the gray triangles indicate the result of evaluating their corresponding expressions.

A minus sign cannot be used to negate a number, only for subtraction. (So 6×-5 is not allowed, but 6×7-5 is.) A number cannot have a leading 0, unless it is actually 0. PEMDAS order of operations applies here.

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  • $\begingroup$ Penpa+ link for interested solvers. $\endgroup$
    – ACB
    Commented Jan 20 at 9:27
  • $\begingroup$ Multiple white fields in a row may be all filled with digits to create a multi-digit number? Or do you have to always alternate digits and operations? $\endgroup$
    – quarague
    Commented Jan 20 at 9:27
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    $\begingroup$ @quarague you can deduce that concatenation of digits must be allowed because for example 5 vertical cannot start with an operator then 102 horizontal must start with the concatenation of at least 2 digits $\endgroup$
    – Ivo
    Commented Jan 20 at 10:09

2 Answers 2

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Some general observations:

• The three-cell expressions are single-digit operator single-digit.
• An expression cannot start with a zero or with an operator. The top row and the left column must therefore be three-digit-number operator three-digit-number.
• The fractions must involve division, but we don't know the operands yet.
• Because 5 in the NE corner must be obtained by division, the expression must be 5 ÷ 1.
• The prime numbers 11 (centre) and 13 (SW) can only be achieved by addition.
• Because 18 (SW) must be obtained by addition, the expression is 9 + 9.
• The 42 (NE) can only by achieved by multiplication.

first step

The left column and the top row:

• The long expression fpr 47047 in the left column must be a multiplication of three-digit numbers: 47047 = 329 × 143.
• Either the 2 or the 4 must be the first operand to 13 (SW). The operation is addition, so the expression must be 4 + 9.
• The expression to 2/3 (NW) is now 2 ÷ 3.

• the operation for 102 (top row) can be a subtraction.
• the expression starts with "35". The middle digit of the second operand must be a divisor of 8, so one of 2, 4 or 8. We have a subtraction with carry, so it is 4.
• Because the down expressions cannot start with a zero, the expression must be 351 − 249.
• The expression for 16 (NE) is now 4 × 4.

halfway there

More progress

• The expression for 94640 must be a multiplication of a two-digit and a four-digit number. The first two digits of the expression are "91" and conveniently, 94640 = 91 × 1040. (The expression might still be something else, but let's got with that for now, especially since the 4 also fits.)
• That means that 8 (centre) is 1 + 7. •The expession for 56/3 (18 2/3, centre bottom) could be a division of a four-digit number 19xx and a two-digit number. The fraction is already simplified, so the numerator should be a multiple of 56 to expand the fraction. Unfortunately, expanding with 34 and 35 leads to the three-digit denominators 1904/102 and 1960/105. • Alternatively, the expression can be 19 − 7 ÷ 21. (The seven was already there and the denominator can't be i the tens, because the tens digit must be the nuerator for 2.)
• That gives us 2 = 2 ÷ 1 (SE).
• The right column is a division and we know that the tens digit of the first nuber is either 6 or 7. It can't be 7, because 970 isn't divisible by 8, so it is 8 == 960 ÷ 120.
• The expressions for 6 (NE) and 1/2 (SE) now fall into place.

nearly there

Last steps and the final grid:

• The expression for 1 (third column) is 13 × 7 − 90.
• 1000 == 390 + 610 (bottom row).
• 169 = 2704 ÷ 16 (fifth column).

final grid

(I hope the write-up makes sense. When writing this answer, I realized that in my original solve, I did some things in a different order.)

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    $\begingroup$ Guess we were both writing at the same time. But you beat me by 2 minutes. :) You have also explained it much better that mine. $\endgroup$
    – ACB
    Commented Jan 20 at 10:29
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    $\begingroup$ Yes, a close call, but with such long answers, two minutes is nothing. I'm also not sure whether my explanations are better. I like how you highlight the operators and show possible candidates. My explanations are quite wordy. Why not un-delete your answer? I think it has additional value to mine. $\endgroup$
    – M Oehm
    Commented Jan 20 at 10:42
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    $\begingroup$ By the way, if you'd like to use true list formatting instead of Unicode bullet points, HTML works in spoilers. $\endgroup$
    – bobble
    Commented Jan 20 at 15:34
  • $\begingroup$ Minor correction in the first spoiler: "Because 5 in the NW corner ..." $\endgroup$
    – Christoph
    Commented Jan 22 at 11:06
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Answer

Answer

Solve path

We can easily see that blue-shaded cells should contain arithmetic operators. Moreover we can say that R2C2 and R7C7 are ÷s.

1

With that we can mark 5÷1 vertically in C2.

Next focus on the 13 in R7. It should be made with two digits. The only operation to do this is addition. Therefore we can mark a + in R7C2, then 9+9 in the bottom half of C2.

With a similar logic we can put + in the middle cell (considering 11) and × in R2C7 (considering 42).

2

To make 47047, multiplication is a must. Looking at the factors of 47047, we can find the only fit 329×143.

Then we can also fill 13=4+9.

3

Considering C3, we see that operators should be placed in either R3 or R5. But we can immediately see that both cells should contain operators to get 1.

4

Now we can complete the row for 94640. Note that it should be of the type 2-digits × 4-digits (Hint: 91×99×9).

After that, we can put a ÷ in R5C5.

5

In R5, note that 56/3=18+2/3 which can be written as 19-1/3.

6

For C5, there are only a few candidates to consider.

2704÷16
4901÷29
5408÷32
7605÷45

We can see that the first option is the perfect fit. The rest follows easily.

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