1.what is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers and the sum of eleven consecutive integers?
2.three different non-zero digits are used to form six different 3-digit numbers, the sum of five of them is 3231.what is the sixth number?
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< 31 sum of six digits < 39
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1.what is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers and the sum of eleven consecutive integers?
I didn't read the question carefully, sorry about the error.
the correct one should be 9a=10b+5=11c
where a is the middle of the 9 numbers and b is the 5th number of the 10 numbers, c is the middle of the 11 numbers.
thus this number should be a common multiple of 11, 9, 5,so it is 5*9*11=495
2.three different non-zero digits are used to form six different 3-digit numbers, the sum of five of them is 3231.what is the sixth number?
let the three digits be a, b, c, then the sum of six numbers is 200*(a+b+c)+20*(a+b+c)+2*(a+b+c)=222*(a+b+c)
222*(a+b+c)=3231+the 6th number.
we know the 6th number is between 100 to 1000, so the a+b+c is between 15 and 19,
1）a+b+c=15, the 6th number is 99, can't be
2) a+b+c=16, the 6th number is 321, but the sum of the 3 digits is not 16, so can't be
3) a+b+c=17, the 6th number is 543, the sum of the 3 digits is 12, not equal to 17, can't be
4) a+b+c=18, the 6th number is 765, the sum of the 3 digits is 18
5）a+b+c=19,the 6th number is 987, the sum of the 3 digits is not 19
so the 6th number is 765
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Q1, the answer is 495. It is a number which is a multiple of 9, 5, and 11
9a_5=11(b_5+b_6)=11C_6
sum from 51 to 59
from 45 to 54
from 40 to 50
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真是高人!辛苦了!非常感谢帮忙!
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第一题可以列方程为：
x + x + 1 + ... + x + 8 = y + y + 1 + ... + y + 9 = z + z + 1 + ... + z + 10
然后推导出：
9x + 36 = 10y + 45 = 11z + 55
得：
x = 10/9 y + 1
y = 11/10 z + 1
即可得z为10得倍数而y为9的倍数。
所以最小的符合值为：z = 40, y = 45, x = 51。
则第一题的解为：495
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第二题可以建模为：
设三个非零整数位x, y, z，则所有的六个数为：
100x + 10y + z
100x + 10z + y
100y + 10x + z
100y + 10z + x
100z + 10x + y
100z + 10y + x
则可以列方程：
222 (x + y + z) = 3231 + m
m = 100i + 10j + k {i, j, k 为x ,y, z的某种组合}
最终试得的结果为：
222 * (7 + 6 + 5) - 3231 = 765
即m = 765
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