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Math problems in grade one, about finding rules, urgent need,
1 .10 times 10:

Formula: head joint, tail to tail, tail to tail.

For example: 12× 14=?

Solution: 1× 1= 1.

2+4=6

2×4=8

12× 14= 168

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

2. The heads are the same and the tails are complementary (the sum of the tails is equal to 10):

Formula: After a head is added with 1, the head is multiplied by the head and the tail is multiplied by the tail.

For example: 23×27=?

Solution: 2+ 1 = 3

2×3=6

3×7=2 1

23×27=62 1

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

3. The first multiplier is complementary and the other multiplier is a number.

Same:

Formula: After a head is added with 1, the head is multiplied by the head and the tail is multiplied by the tail.

For example: 37×44=?

Solution: 3+ 1=4

4×4= 16

7×4=28

37×44= 1628

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

4. Eleven times eleven:

Formula: head joint, head joint, tail to tail.

For example: 2 1×4 1=?

Answer: 2×4=8

2+4=6

1× 1= 1

2 1×4 1=86 1

5. 1 1 times any number:

Formula: head and tail do not move down, middle and pull down.

For example: 1 1×23 125=?

Answer: 2+3=5

3+ 1=4

1+2=3

2+5=7

2 and 5 are at the beginning and end respectively.

1 1×23 125=254375

Note: If you add up to ten, you will get one.

6. Multiply a dozen by any number:

Formula: The first digit of the second multiplier is fixed.

Downward, the unit of the first factor is multiplied by the second factor.

Each number, add the next number, and then drop.

For example: 13×326=?

Solution: 13 bit is 3.

3×3+2= 1 1

3×2+6= 12

3×6= 18

13×326=4238

Note: If you add up to ten, you will get one.