Change the operation order of 671 + 29 × 4 ÷ 2 to sum first, then to product, and finally to quotient______ ．

Change the operation order of 671 + 29 × 4 ÷ 2 to sum first, then to product, and finally to quotient______ ．

The operation order of 671 + 29 × 4 △ 2 is changed to sum first, then product, and the final quotient should be: (671 + 29) × 4 △ 2, so the answer is: (671 + 29) × 4 △ 2

4 / 7 times 29 + 42 times 4 / 7-4 / 7 (simple operation)

=4/7(29+42-1)
=4/7*70
=40

1 2 4 8 16 32 64 128 256 512 1024... How can you calculate the sum of any number of these numbers?
1 + 4 + 8 = 13, how to calculate the sum of the three numbers 1, 4 and 8?
4 + 32 + 64 = 100, how to calculate the sum of the three numbers 4, 32 and 64?
How do you add these four numbers 2 + 512 + 8 = 650?
Any number can be added by one, two, three, four, etc. (each number can only be used once without repetition)
How to calculate a number 172 directly is the result of adding those numbers (4 + 8 + 32 + 128 = 172)

This is a binary problem
Like 13
13÷2=6…… 1 & nbsp; (the last bit of binary is 1)
6 △ 2 = 3 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (the penultimate bit of binary is 0)
3÷2=1…… 1 & nbsp; & nbsp; (the penultimate bit of binary is 1)
The last quotient is 1, so the highest bit of binary is 1
So the binary representation of 13 is
one thousand one hundred and one
So, select (from the right) the first, third, and fourth numbers, that is, 1, 4, and 8
The binary representation of 172 is
10101100 (imitate my previous introduction)
So select (from the right) the third, fourth, sixth and eighth numbers, that is, 4, 8, 32 and 128