One key thing to note here is that we do not need to consider the 11 paise coin, as the 11 paise coin can always be expressed as one 4 paise and one 7 paise coin.
So, our question boils down to finding the largest amount that we cannot pay using only 4 and 7 paise coins.
Since these are relatively small values, we can try some of the initial values until we reach the point where four consecutive numbers can be expressed, as every group of 4 numbers after that can be expressed using the same coins with an additional 4 paise coin.
A faster and more direct approach would be to use the Chicken McNugget theorem, also known as the Frobenius coin problem or the postage stamp problem, which essentially says that for any two relatively prime numbers, m and n, the greatest integer that cannot be written in the form am+bn is given by $$mn-m-n$$
Here, since we eliminated 11, and only need to deal with 4 and 7 paisa coins, we can use the formula to get the highest amount that cannot be paid using these two values to be $$28-4-7=17$$
Alternatively:
4 = 0(mod 4)
7 = 3 (mod 4)
14 = 2(mod 4) and
21 = 1(mod 4)
Any amount greater than 21 paisa, can be paid using these three coins.
For example, if the amount leaves a remainder of 0 when divided by 4, we can pay using 4 paisa coins only.
If it leaves a remainder of 1 when divided by 4, we can pay three 7 paisa coins and the rest by 4 paisa coins.
Similarly if it leaves a remainder of 2 when divided by 4, we can pay two 7 paisa coins and the rest by 4 paisa coins.
Similarly if it leaves a remainder of 3 when divided by 4, we can pay one 7 paisa coin and the rest by 4 paisa coins.
Of the numbers less than 21 paisa, the maximum we can't pay using these coins is 17 paisa coins.
Hence, answer is 17 paisa.
