At a roadside stand, Paulo sells small jars of honey for $4

Question

At a roadside stand, Paulo sells small jars of honey for $4 and large jars for $7. How many two-digit integers cannot be his revenue from the sale of jars of honey yesterday?

(A) 2
(B) 3
(C) 4
(D) 5
(E) More than 5

HarshavardhanR · Answer

Small Jar 
 Large Jar 
 
  Selling Price 
 $4 
 $7 
 
  #Units  
 a 
 b

Question:  How many 2-digit integers (10 to 99) cannot be the revenue from sale of honey jars?

Revenue from sales of honey jars yesterday =  4a + 7b

There are only 4 types of integers between   -> 4k, 4k + 1, 4k + 2, and 4k + 3.

(1) 4k -> 12, 16, 20, ....

Clearly, all these 2-digit integers are possible revenue values. (4a + 7b) with b= 0 will give us all the above values.

(2) 4k + 1 -> 13, 17, 21, 25, 29, ...

- Can 13 be written as 4a + 7b? No. 
- Can 17 be written as 4a + 7b? No. Remember -> given 4b is even, for 4a + 7b to be odd, 7b has to be odd too. b= 1,3, etc. Impossible to write 17 as 4a+7b
- Can 21 be written as 4a + 7b? Yes. 4(0) + 7(3)

Observe:  From 21 onwards, we don't need to check further. Because 21 = 4(0) + 7(3). Every further number in this series can be obtained by just adding 4s.

-> 25 = 4(1) + 7(3)
-> 29 = 4(2) + 7(3)
-> Etc.

Key insight:  Once we hit a 7-multiple (like 21), that and every subsequent number can easily be expressed as 4a+7b. We won't need to worry about those numbers!!

So far ->  we have identified 2 2-digit integers that cannot be the revenue .

(3) 4k + 2 -> 10, 14, 18, 22, ...

- Can 10 be written as 4a + 7b? No.
- Can 14 be written as 4a + 7b? Yes.

So, every subsequent number in this series (18, 22, etc.) can easily be expressed as 4a+7b.

-> 18 = 4(1) + 7(2)
-> 22 = 4(2) + 7(2)
-> 26 = 4(3) + 7(2)
-> Etc.

So, we don't need to worry about any of these numbers.

So far ->  we have identified 3 2-digit integers that cannot be the revenue . (10, 13, and 17)

(4) 4k + 3 -> 11, 15, 19, 23, 27, 31, 35, 39, 43, ...
  
Given that 7 itself is 4k+3,

4K+3 -> 7, 11, 15, 19, ......

Of course, we are concerned about 2-digit numbers only. But we can clearly see that every single number in this series can be expressed as 4a+7b.

7 -> 4(0) + 7(1). And so on. If we forget about "7" being 4k+3, that is not a problem. We could still check carefully from the 1st 2-digit 4k+3, till we hit a 7-multiple, as we have done in the earlier cases.

-> 11 = 4(1) + 7(1)
-> 15 = 4(2) + 7(1)
-> 19 = 4(3) + 7(1)
-> 23 = 4(4) + 7(1)
-> 27 = 4(5) + 7(1)
-> 31 = 4(6) + 7(1)
 -> 35 = 4(0) + 7(5)
 
-> 39 = 4(1) + 7(5)
-> 43 = 4(2) + 7(5)
-> Etc.

In any case, we can clearly see that among the numbers of the form 4k + 3, every single number can be expressed as 4a + 7b.

So,

Overall->  There are exactly three 2-digit integers that cannot be the revenue . (10, 13, and 17).

---
Harsha

kevincan · Answer

What a wonderful explanation!

kevincan · Answer

10
11=7+4
12=4*3
13
14=2*7
15=7+2*4
16=4*4
17
18=7*2+4
At this point onwards, every greater positive integer is a multiple of 4 greater than either 0,7,14 or 21, as these numbers leave a remainder of 0,3,2 and 1 when they are divided by 4.

At a roadside stand, Paulo sells small jars of honey for $4

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At a roadside stand, Paulo sells small jars of honey for $4

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