brigittew

67、还有那道abcd都为质数，问除了1和自己，有多少个不同的因子。
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answer: 16

the answer here is wrong
as we know, we calculate the number of different factors, the equation is (n1=1)*(n2=1)...so here should be (1+1)*(1+1)*(1+1)*(1+1)=16. since here we need exclude 1 and itself, so here left 14.
i once met such kind of questions in my simulation test, just as, there are how many factors of 196 is the multible of 6
firstly, we need calculate the number of different factors of 192, then further account the exact number of the multiple of 6

brigittew

7.对于正整数N，N前面的数包括N的和是N（N+1）/2。请问714前包括714的所有6的倍数的和是多少？
[讨论]3*119*120
we can get the number of multiple of 6 here from n=714, it is 119
why answer here isnt 119*(119+1)/2
since'对于正整数N，N前面的数包括N的和是N（N+1）/2' here N refers to the total number instead of the exact number such as 6(though here the question is to ask the sum of the all numbers of multiple of 6)

brigittew

120.140N是一个整数的平方，问N最可能是那个数？选项有35。。。
[讨论]有问题，140Ｎ什么意思？相乘？

here is my way:
140*n=y^2
2^2*7*5*n=y^2 since we need get a y^2here, n at least equal to 35