(2) Use these five numbers to form another five numbers at will.
(3) Subtract two five-digit numbers (reduce large numbers).
(4) Let the other person think of any number and tell you other numbers (except the one that the other person thinks of).
5. The performer only needs to add the number spoken by the other person to one digit and subtract 9 to know what the other person is thinking.
Example: five digits one: 57429; Five digits two: 24957; Subtraction: 32472;
Remember in your heart: 7; Tell the rest to the executor: 3242;
Performer: 3+2+4+2 =11; 1+ 1=2; 9-2 = 7 (that is, the number remembered by the other party)
I thought it was interesting, and then I proved it.
When I was in elementary school, I said that if an integer is a multiple of 9, then only if the sum of its digits is also a multiple of 9. A little generalization, if the remainder of an integer divisible by 9 is R, then if and only if the remainder of its digits divisible by 9 is R .. first prove it.
Necessity:
n = d 1+ 10×D2+ 100×D3......=σ( 10i×di)( 1)
9k + r = d 1 + d2 + d3......=σdi(2)
( 1)-(2)
n-9k-r =σ(( 10i- 1- 1)×di)，n =σ(( 10i- 1- 1)×di)+9k+r
∫ 10i- 1 =( 10- 1)( 10i-2+ 10i-3+ 10i-4+....)
∴ n = 9× (σ (10i-2× di)+k)+r, that is, the remainder of n divisible by 9 is R.
Adequacy:
n = 9k+r = d 1+ 10×D2+ 100×D3......=σ( 10i×di)
9k+r =σdi+σ(( 10i- 1- 1)×di)
σdi = 9k+σ(( 10i- 1- 1)×di)+r = 9k+9×σ( 10i-2×di)+r = 9(σ( 10i-2×di)+k)+r
That is, the remainder of σ di divisible by 9 is R.
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Then it is proved that the difference between any two decimal numbers with the same number of digits is a multiple of 9.
∫ Two numbers, all the numbers are the same,
The sum of these figures is the same,
The sum of these numbers is the same as the remainder of 9.
The remainder of these two numbers divisible by 9 is the same.
The difference between these two numbers is a multiple of 9.
The difference between two multiples of 9 is also a multiple of 9.
□
After removing a number, the sum of digits is 9k-d, and the remainder divisible by 9 is 9-D. ..
The sum of numbers is 9k+(9-d)
The remainder of the sum of numbers divisible by 9 is also 9-d.
The remainder of the final result divisible by 9 is also 9-D.
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In fact, it can be seen from the proof process that no matter how many digits there are, it is not necessary to have five digits.