Use 1, 2, 3, 4 and 5 to form a 2-digit number and a 3-digit number. What are the maximum and minimum product numbers? How can we calculate it as soon as possible

Use 1, 2, 3, 4 and 5 to form a 2-digit number and a 3-digit number. What are the maximum and minimum product numbers? How can we calculate it as soon as possible

#include
using namespace std;
int main()
{
int a,b,ans,max,min;
for(int i=1;i
Maximum: 531 * 42 = 22302
At the maximum, two multipliers should be as large as possible, from high to low, 5 rows of 3-digit hundred, 4 rows of 2-digit ten, 3 rows of 3-digit ten
Minimum: 135 * 24 = 3240
At the minimum, the two multipliers should be as small as possible, from low to high, 1 row of 3-digit hundred, 2 rows of 2-digit ten... How can we calculate it as soon as possible? The maximum: 531 * 42 = 22302, when the maximum, the two multipliers should be as large as possible, from high to low, 5 rows of 3-digit hundred, 4 rows of 2 tens
Maximum: 531 * 42 = 22302
At the maximum, two multipliers should be as large as possible, from high to low, 5 rows of 3-digit hundred, 4 rows of 2-digit ten, 3 rows of 3-digit ten
Minimum: 135 * 24 = 3240
At the minimum, the two multipliers should be as small as possible, from the low to the high, the first row is 3-digit hundred, the second row is 2-digit ten
Xiao Ming read a story book. On the first day, he read 40% of the whole book. On the second day, he read 60 pages. There are 48 pages left. How many pages are there in this book?
Suppose there are x pages in the whole book
40%X+60+48=X
We can get x = 180
There are 180 pages in the book
Calculation: 1 + 2-3 + 4 + 5-6 + +2005+2006-2007=
0+3+6+…… +2004=3（1+2+…… +668）=3*669*334=1002*669=670338
（1+2-3）+（4+5-6）+......+（2005+2006-2007）
=0+3+......+2004
=（0+2004）*668/2
=2004*334
=669336
The first group: 1 + 2-3, the second group: 4 + 5-6, the third group A total of 2007 numbers, can be divided into 2007 △ 3 = 669 numbers, each group is 0 + 3 × 1 + 3 × 2 + +3×（669-1）＝3(1+2+3+… ... + 668) the rest will count~~
1+2-3+4+5-6+…… +2005+2006-2007=0+3+6+9+......+2004=3(1+2+3+....+664)=662340
（1+2007）x（2007÷2）=670338
If the product of 2 boxes multiplied by 4 is two digits, what is the maximum number in the box? If the product is three digits, what is the minimum number in the box?
24×4=96
25×5=100
Fill in 4 and fill in 5
4 5
I'm wrong. I can fill in 2 at most
In the second day, there were 43 pages of stories in the book. How many pages did you read in the first day?
(42 + 43) △ 1-83%, = 85 △ 17%, = 500 (pages); a: this story book has 500 pages
Simple calculation of sixth grade math problems 2006 * 2008 * (1 / 2006 * 2007 + 1 / 2007 * 2008)
The correct answer will have a high reward, please write the calculation process
2006 * 2008 * (1 / 2006 * 2007 + 1 / 2007 * 2008)
=2008/2007+2006/2007
=2
Original formula = 2006 * 2008 * (1 / 2006 minus 1 / 2007 plus 1 / 2007 minus 1 / 2008)
=2006 * 2008 * (1 / 2006 minus 1 / 2008)
=Ask: No, I wrote the formula right
Use 1, 2, 3, 4, 5 to form a three digit number and a two digit number to maximize the product of the two multiplication
431*52=22412
431*52=22412
How many pages did Xiao Ming read in the first day?
Please choose one of the following conditions to fill in the brackets and answer it
① There are 132 pages in two days. There are 48 pages left
③ There are still four fifths of the book left to be read. 4. The ratio of pages read to pages not read is 11:4
(1) (132-60)/40%=180
(2) (48+60)/(1-40%)=180
(3) 60/(1-40%-4/15)=180
(4) 60/(1-40%-4/(11+4))=180
1. The suburban vegetable base transported a batch of vegetables to the city. The original plan was to load 1 200 kg per car, but actually 1 680 kg per car. As a result, two cars were used less. How many cars were planned?
2. Two engineering teams build the same road. Team a builds 168 kilometers from east to west every day, and team B builds from west to East. After five days, the two teams meet. It is known that team a needs 15 days to build the road alone, but how many meters does team B build every day?
1. 1680 * 2 / (1680-1200) = 7 (vehicles)
Answer: originally planned 7
2. 168 * 15 / 5-168 = 336 (km)
A: Team B repairs 336 meters a day
1. (1680-1200) △ 2 = 240 (vehicles)
A: the original plan is to use 240 cars.
2.168 ÷ (15-5) = 16.8 (m)
A: the team has repaired 16.8 meters a day.
1. (1680-1200) divided by 2 = 240
2.168 divided by (15-5) = 16.8
1. 1200 * x = 1680 * (X-2) x = 7 vehicles
2. (168 + x) * 5 = 168 * 15 x = 336km
1, 2, 3, 4, 5 and 6 make up two three digit numbers to maximize their product
631 × 542 = 342002 is the largest!
First of all, one of these two numbers must start with 6 and the other with 5, otherwise the number is obviously small
There are six combinations at the beginning of 6 and 5, which are written according to the way of making up the largest number. We can get the following six formulas
621×543=337203
631×542=342002
641×532=341012
632×541=341912
642×531=340902
643×521=335003
Obviously 631 × 542 = 342002 is the biggest!
Five and six must be hundreds
Three and four must be ten
One and two must be a bit
According to experience, it should be 631 542
Do not repeat! If so, it is:
Obviously, the first two numbers are 6 and 5 respectively. For other numbers, they are all smaller. For example, 65A * 4bc < 660 * 440 = 2904 < 600 * 500;
Let the two numbers be 6ab and 5CD respectively, then C must be 4. Suppose a is 4, then:
64b*53d=640*530+530b+640d+bd=630*530+530（b+d)+110d+bd+5300；（1）
63b * 54d = 630 * 540 + 540B... Expansion
Do not repeat! If so, it is:
Obviously, the first two numbers are 6 and 5 respectively. For other numbers, they are all smaller. For example, 65A * 4bc < 660 * 440 = 2904 < 600 * 500;
Let the two numbers be 6ab and 5CD respectively, then C must be 4. Suppose a is 4, then:
64b*53d=640*530+530b+640d+bd=630*530+530（b+d)+110d+bd+5300；（1）
63b*54d=630*540+540b+630d+bd=630*530+530（b+d)+100d+10b+bd+6300；（2）
(2) - (1) we get: 1000 + 10b-10d > 900 > 0, the first two combinations should be: 63 and 54;
631 * 542-632 * 541 = 631 * 541 + 631-632 * 541 = 631-541 = 90, that is, the combination of this number is 631 and 542