This is how I solved them.

C, imo. A and B doen't work. The largest 4-digit multiple of 10, 11, 15, and 22 is 9900. Correct answer should be less than 9900.

I think you copied the question wrongly - "no of four digits" is ambiguous. It is probably "four digit number".


Answer should be 4.5


A.
1.Find the LCM of 2,3,4,-9,6,18,36,60. (i.e. 2*2*3*3*5=180)
2. Since we are after the "smallest square number", you need to make sure that each prime factor of 180 has even power. You need to multiply 180 by 5 to achieve this. 900.

What is the source of these questions?

i sincerely hope that you are the only one who considers question 1 and 3 as basic quant.
seofah crazy skills. +1

Hello! Hope I can be of some help.

The first thing I want to ask is, yeah, where are these questions coming from? They have some fundamental flaws in them.

Number 1: Yes, C is correct.
Here's another explanation; seofah's answer quickly and efficiently kills A and B, but it's that D that makes me nervous. We begin.

A clearly fails statement 1: in order to have a remainder of 3 over 10, it must end in 3. FAIL!

B passes statement 1 (it ends in 3), but then fails statement 2. Explanation follows.

Here's a fun trick: To see what remainder a number would have over 11, alternately subtract and then add the digits. Thus, 9903 --> - 9 + 9 - 0 + 3 = 3. We wanted a remainder of 4. Too bad. FAIL!

(Of course, you can also note that 9903 = 9900 + 3, in other words, 3 more than an obvious multiple of 11.)

C passes statement 1 quite easily. It also passes statement 2. Let me talk about that a little bit more:

Method I : 9893 --> -9 + 8 - 9 + 3 = -1 -6 = -7
Now, you can't really get a remainder of -7; when you divide by 11, you can only get remainders from 0 through 10. Thus, all you do is add or subtract 11s until you get within that range. Thus, -7 + 11 = 4. Bingo. PASS!

Method II: 9893 = 9900 - 7. Therefore, it's 7 less than a multiple of 11, and therefore it's 4 more than the previous multiple of 11, which by the way happens to be 9889. PASS!

Now to statement 3: a remainder of 8 over 15. Hmm. Oh! Cool!
If that's the case, then the number 9893 must be 8 more than a multiple of 15, right? Thus we turn our attention to the number that's 8 less, namely 9885. In order to be a multiple of 15, it must be a multiple of both 3 and 5 (this is true whenever the divisor you're testing has factors that are relatively prime*). Clearly it passes the 5 test (ending in 5 or 0). Do you know the test for 3? You add the digits, and if that result is divisible by 3, then so was the original. 9885 --> 9 + 8 + 8 + 5 = 30, which is clearly divisible by 3. PASS.

And we'll use something similar to test the remainder of 15 over 22. 15 less gives us the number 9893 - 15 = 9878. Divisibility by 22 requires it to be divisible by 2, which... ok PASS now we look at 11.
9878 --> -9 + 8 - 7 + 8 = -1 + 1 = 0
And a remainder of 0 over 11 means that it is actually divisible.

QE freakin D, y'all. Thanks for the workout! That was fun. I'll be back later to tackle the others.

TAKE AWAY POINTS:

* When dividing by N, the only legal remainders lie within the range 0 through N-1.
* If a test gives you remainder outside that range, then add or subtract Ns until you get within that range. Those familiar with the Chinese Remainder Theorem (seofah, sounds like you might...) will know what I'm talking about.
* The 3 test: Add up the digits of N. This result has the same remainder over 3 as N did. Therefore, you can repeat this procedure.
* The 11 test: Do this on the digits: - + - + (remember to start with minus!) This result has the same remainder over 11 as N did to begin with.
* Relatively prime: having no factors in common. Thus, the test for 6 is just the test for 3 as well as the test for 2. However, you can't test 12 by testing 2 and 6 (think about why this is stupid... anything that passes 6 would automatically pass 12!). To test for 12, you should test 3 and 4. The 33 test is the test for 11 and the test for 3, and so on and so forth. This allows you to create massive hybrid tests.

Examples:

3 test:
415 --> 4 + 1 + 5 = 9, which is evenly divisible by 3, and so 415 was as well.
5135 --> 5 + 1 + 3 + 5 = 14 --> 1 + 4 = 5, which has a remainder of 2. Thus not divisible.
5135 --> 5 + 1 + 5 = 11 --> 2. When doing the 3 test, you can automatically ignore any 3s, 6s, and 9s that come up.

11 test:
1331 --> -1 +3 -3 +1 = 0. Thus it is divisible (did you recognize it as 11^3? Think Pascal's Triangle.)
12 --> -1 + 2 = 1. By the way, this is the one I use to remember which way to do the signs. You don't want a remainder of -1 which is 10. Obviously.

4567 --> - 4 + 5 - 6 + 7 = 2.
Thus, the number that's 2 less, 4565, can be written as
4400 + 165 = 4400 + 110 + 55. Ah. All 11s.

The funny thing is that it is possible that the OP did NOT copy the question incorrectly; it could work in one sense. "Number, of four digits" made. Like, a house of brick. Now, I ain't gonna lie, I agree that Seofah's restatement is much, much better. Just being goofy