is this confusing or tricky?

Question

1.when the 3 digit numbers 4a3 and 5b2 are added together the answer is divisible by 9. what is the largest possible value of a and b that will satisfy the condition.
2.suppose you were offered a job and were promised one centavo on the first day, 2 centavos on the second day, 4, on the third day and so on with your salary on each succeeding day double that of the previous day for everyday of the month. would you take it? ow figure out how much you would have received had you taken the job for the month of march.
3. the 4 digit number A55B is divisible by 36 without remainder. What is A and B?

Answer 1

1. 4a3 is actually 4*100+a*10+3*1
and 5b2 is 5*100+b*10+2*1

when you add them, u get, 9*100+(a+b)*10+5*1
this number can be written as 4(a+b)5
in order that this number is divisible by 9, the sum of the digits must be a multiple of 9. therefore (4+5+a+b)=(9+a+b) should be divisible by 9. and (a+b) is a single digit so that a and b can take values from 0 to 9 that too such that a+b is less than 10
which gives the following pairs with sums

0,9 = 9
1,8=9
2,7=9
3,6=9
4,5=9
9,9=18
but 18 is a two digit no. so the largest possible value of a and b could be 4,5 or 3,6

2. the number of centavo's are in a geometric progression with common ratio 2
the sequence is 2,4,8,16, so on
this is a very profitable job indeed. for the month of march, the number of days are 31 so sum of such 31 terms will be

as shown, sum= 2*(2^31-1)/(2-1)= 4294967294
note: for a geometric progression with first term a and common difference r, the sum of the first n terms is a*(r^n-1)/(r-1).

4. for A55B to be divisible by 36, it should be divisible by 4 and 9

if it is divisible by 4 then the last two digits have to be 52,56
so b can be 2 or 6

now, if the no. is divisible by 9 then sum of the digits must be div by 9 so that 10+A+B should be div by 9. thus 10+A+2 or 10+A+6 should be a multiple of 9, A being a no. from 0 to 9

so A has to be either, 6 if B is 2, or 2 if B is 6
so A is 2, B is 6
or A is 6 and B is 2

phew, that took a lot of typing.... :-)

Answer 2

1. Not really, no. If you're aware that the digit sum of any number divisible by nine is also divisible by nine, this one's pretty simple. Just find a way to express the sum of 4a3 and 5b2, then add together that number's digits. The result should be 18, and you can work backwards from there.
2. I would LOVE if I received this offer. This guy is trying to make you rich. The money you have on the nth day is (2^n)-1, which is enormous by the end of the month.
3. This one will be a little similar to #1, but it WILL be a little more tricky. If you don't have a satisfactory answer by this afternoon, I'll come back and explain more.

Answer 3

a can be maximum 9 , b nine

a+b <= 18

now 4a3+5b2 is divisible by 9 so 4+a+3+5+b+2 divisible by 9
so 14+ a+b divisible by 9

a+b = 4 or 13

the comination can be worked out

2) very large 2^30-1 on a mormal month and 2^31-1 in 31 day month

3) A+B = 8 as sum of digits divisible by 9

divisible by 4 so 52 or 56 ending
numbers are 6552 or 2556

Answer 4

Hint for 1: If a number to be divisible by 9, the digits have to add up to 9.

Hint for 2: Exponential functions increase really fast

Hint for 3: A number divisible by 36 is also divisible by 9 and 4. For divisibility by 9, see hint for #1. If it is diviisible by 4, the last two digits must be divisible by 4. That greatly limits the possibilities.

Answer 5

1.
I don't know.



2.
Yes!
March has 31 days. 2^(31 – 1) = 1,073,741,824 centavos!



3.
A = 2
B = 6

Answer 6

Number 2, I got 1073741824. I don't know if it's right though.

Answer 7

Its both confusing and tricky
perhaps easier q's would get more answers