Simple calculation of 17 × 19 + 17

Simple calculation of 17 × 19 + 17

17×19+17
=17×(19+1)
=17×20
=340

97×2000-96×2001．

97×2000-96×2001，=97×2000-96×（2000+1），=97×2000-96×2000-96，=（97-96）×2000-96，=2000-96，=2000-100+4，=1900+4，=1904．

Simple calculation of 2000 × 2000-1997 × 2001 = 4003

（2001-1）×2000-1997×2001
=（2000-1997）×2001-2000
=6003-2000
=4003

97×2000-96×2001．

97×2000-96×2001，=97×2000-96×（2000+1），=97×2000-96×2000-96，=（97-96）×2000-96，=2000-96，=2000-100+4，=1900+4，=1904．

1996, 1997, 1998, 1999, 2000, 2001, 2002 how to calculate in a simple way?

Add? You can do that
Original formula = 1999 * 4 = (2000-1) * 4 = 8000-4 = 7996

Evaluation 1999 × 2000 × 2000 / 2000 × 1999 × 1999

Write the formula 1999 × 2000 × 2000 / 2000 × 1999 × 1999 into a fraction. The numerator and denominator of 1999 and 2000 are about to be deleted. The numerator is only 2000 and the denominator is only 1999, so the answer is 2000 / 1999
If you don't understand, you can ask me again

Two 100 digits 111 11 (100 ones) and 999 The product of 99 (99 nines) has () odd numbers

11 * 9 = 99111 * 99 = 109891111 * 999 = 110988911111 * 9999 = 111098889, (sequence an is 111 11 (n + 1 one) and 999 99 (n 9) product, n > = 2, then an = 111 (n-1 1) 098 (n-1 8) 9, when n = 99, there are 98 1 and 2 9 in the product, a total of 100 odd numbers

There are several even numbers in the product of 111... 11 (2010 1) × 999.. 99 (2010 9)

1 × 9 = 9 has 0 even numbers;
11 × 99 = 1089 has two even numbers; 0 and 8
111 × 999 = 110889 has 3 even numbers: 0, 8, 8
1111 × 9999 = 11108889 has 4 even numbers; 0, 8, 8, 8
11111 × 99999 = 1111088889 has 5 even numbers; 0, 8, 8, 8
…… ……
According to this rule, the product of 2010 1 × 2010 9 contains 2010 even numbers

555…… (10 5) and 999 What is the sum of the digits of the product of (10 nines)

5

How many odd numbers are there in the product of 111 '` 11 and 999' ` 99?
If you can explain it in detail, it would be better

111```11*999```99=111```11*(100```000-1)=111```1100```000-111```11