f N is any positive integer, how many consecutive integers following N are needed to insure that at least one of the integers is the integers is divisible by another postivie inter m?
A. m-1
B.m
c. m+1
d. 2m
e. m square
how many of three-digit numbers containing no digits other than 2,3, or 4 are divisible by 4?
a. 2
b. 3
c. 4
d. 6
e. 9
Should really be posted in DAT forum. For the second question I've heard this one before but it was divisible by 3, not 4. For divisible by 3, the answer is 9. Divisible by 4 is easier since the division by 4 rule states that the last 2 digits have to be divisible by 4. So with 2, 3, and 4 you could have only 24, 32, or 44. Since each of those could start with one of the 3 digits, that's 3x3=9 (example 24 can be 224, 324, or 424).
For the first one consider an example. Say N = 101 and m = 10. It wants to know how much we need to add to 101 to be SURE that one of those numbers will divide 10 evenly. Turns out we need to add 9 to get 110 which divides 10 evenly. Now that you know what the problem wants, you need to think more generally.
Say you have two numbers N and m. When you divide N by m, it either goes in evenly or it has a remainder. What can the remainder be? Anything between 1 and m-1. If we add 1 to N (have N+1), the remainder will INCREASE by 1. If we keep adding 1 more to that number, the remainder will continue to increase by 1 until we hit m-1. Then if we add 1 more to the number and divide by m, we'll cycle back around to remainder 0 - dividing evenly.
Example: N = 14 and m = 4. We start with 14/4 = 3 remainder 2. Add 1 to N and get 15. 15/4 = 3 remainder 3. Add 1 to N and get 16. 16/4 = 4 remainder 0.
The question asks for any N and any m. So we want to look for the 'worst case scenario' - describe N and m such that you need to add the MOST to N to loop back around to remainder 0. Obviously this would be where N/m gives you a remainder of 1. If N/m divides evenly then there is no remainder and you don't need to add anything to N. But if N/m gives remainder 1, then you have to add the highest possible number to loop back to remainder 0. What is this number? Remember that m-1 is the highest remainder and 1 is the lowest. So (m-1) - 1 = (m-2). You need to add m-2 to get from remainder 1 to the highest possible remainder. By adding 1 more at that point, you cycle back around to remainder = 0. So (m-2) + 1 = (m-1). By considering the (m-1) numbers following N, you are ENSURING that one of those (m-1) numbers OR N ITSELF will divide evenly by m.
Examples:
N = 101, m = 10. You need to consider the next 9 numbers. The 9th one works: 110/10 = 11.
N = 14, m = 4. You need to consider the next 3 numbers. The 2nd one works: 16/4 = 4.
N = 21, m = 7. You need to consider the next 6 numbers. 21 divides 7 already, 21/7 = 3.
N = 205, m = 204. You need to consider the next 203 numbers. The 203rd one works: 408/204 = 2.
Get it?
