Problem description:
(1) What is the natural number when a number is divisible by 5 and 3, 6 and 4, 7 and 5? (2) 1 is divided by 3, 4 and 5. How many numbers are there from 1 to 1000? (3) In a container full of water, pour out half of the water, then pour out one third of the remaining water, continue this method, and pour out one quarter of the remaining water for the third time. By the way, how many times have you poured water, and the water left is exactly one tenth of the original? (4) 1997 In the natural number 1, 2,3 ...1997, you can take at most several numbers, so that the sum of any two of these numbers can be divisible by 50. (5) Pour a third cup of salt water and fill it with water, then pour a third cup of salt water and fill it with water ... * * * 5 times in a row. At this time, the salt in the brine is 0.64g, so how many grams of salt does the poured brine contain? (6) The sum of two natural numbers is 9335. After removing ten digits and one digit, one of them is exactly equal to one tenth of the other. What's the difference between these two figures? My grandfather is six times as old as Xiao Ming this year, five times as old as Xiao Ming in a few years, and four times as old as Xiao Ming in a few years. How old is my grandfather this year?
Analysis:
(1) What is the natural number when a number is divisible by 5 and 3, 6 and 4, 7 and 5?
A: In fact, a number can be divided evenly by 5, 6 and 7 after adding 2. Because the least common multiple of 5, 6 and 7 is 2 10. So this number is 208.
(2) 1 is divided by 3, 4 and 5. How many numbers are there from 1 to 1000?
A: In fact, how many numbers from 0 to 999 are divisible by 3, 4 and 5?
999/3=333 999/4=249 999/5= 199
999/(3 * 4) = 83 999/(4 * 5) = 49 999/(5 * 3) = 66 (integer part only)
999/(3*4*5)= 16
So there are 333+249+199-83 (multiples of 3 and 4 are calculated twice)-49 (multiples of 5 and 4 are calculated twice)-66 (multiples of 3 and 5 are calculated twice)+16.
3) In a container full of water, pour out half of the water, and then pour out the remaining one third. Continue this method and pour out a quarter of the remaining water for the third time. After asking, I poured water several times, and the remaining water was exactly one tenth of the original.
A: The first time was 1/2, the second time was (1/2) * (1-1/3) =1/3, and the third time was1/3 * (.
And so on, to the ninth time, there is110.
4) Among the natural numbers of 1997, 1, 2, 3... 1997 can take several numbers at most, so that the sum of any two of these numbers can be divisible by 50.
Answer: Because the remainder of any number divided by 50 is 0-49, and the sum of any two of these numbers is required to be divisible by 50, these numbers are either integer multiples of 50 or the remainder divided by 50 is 25. (Because the sum of two numbers divided by the remainder of A equals the sum of two numbers divided by the remainder of A and then divided by the remainder of A, that is, two numbers can be divisible by A, then the sum of two numbers divided by the remainder of A can be divisible by A)
1997/50 = 39 (integer part only)
1997/25-39 = 40 (because the integer multiple of 50 should be removed)
So there are 40 at most.
(5) Pour a third cup of salt water and fill it with water, then pour a third cup of salt water and fill it with water ... * * * 5 times in a row. At this time, the salt in the brine is 0.64g, so how many grams of salt does the poured brine contain?
Answer: Because the proportion of salt in the solution is unchanged, the salt in the water is 2/3 for the first time, and 2/3 * (1-1/3) = 4/9 = (2/3) for the second time, and so on, and the fifth time is still (2/3).
So 0.64/(32/243) = 4.86 g.
(6) The sum of two natural numbers is 9335. After removing ten digits and one digit, one of them is exactly equal to one tenth of the other. What's the difference between these two figures?
A: Let this number be1000a+100b+10c+d, then the other number is 100a+ 10b.
Then1000a+100 (a+b)+10 (b+c)+d = 9335.
Because abcd is an integer, it is evaluated accordingly.
Because the unit number is only d, then d=5.
Because there is a in thousands, there may be a carry, so it is discussed that if a = 9, then a+b has a carry, so A is not 9 and a=8.
Therefore, if the value of a d is brought in, it is 100b+ 10(b+c)=530.
Similarly, if B is 5 and the end of b+c is 3, there must be a carry, so B is not 5, B = 4 and C = 9.
So these two numbers are 8495 and 840.
7) My grandfather is six times as old as Xiao Ming this year, five times as old as Xiao Ming in a few years, and four times as old as Xiao Ming in a few years. How old is my grandfather this year?
Answer: suppose grandpa's age is 6x and Xiaoming's age is X.
If there is 6x+a=5(x+a), there is1/4x = A.
6x+a+b=4(x+a+b) 5/ 12x=b
Because a b is an integer, A should be a multiple of 12, and grandfather's age should not exceed 100, that is, 6x≤ 100.
So x = 12.
So my grandfather's age is 12 * 6 = 72.