Coin denominations

Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?

A. 66
B. 67
C. 68
D. 69
E. 70

Could anyone please take a look at this and explain what the q means by "how many different sums from 1¢ to 70¢ can he make with a combination of his coins?"..

The prompt is asking how many different SUMS OF MONEY between 1¢ and 70¢, inclusive, can be formed using up to ten 2¢ coins and ten 5¢ coins.
For example, it is possible to form a sum of 16¢ by combining three 2¢ coins and two 5¢ coins:
(3*2) + (2*5) = 16¢.

From 1 to 70, there are 70 possible sums.
The smallest answer choice is 66.
Implication:
At most, 4 of the 70 possible sums cannot be made from the given coins.
Impossible sums are likely to be VERY SMALL or VERY LARGE.

Small sums:
Of the sums between 1 and 10, the following cannot be formed from 2¢ and 5¢ coins:
1 and 3.

Large sums:
If all 20 coins are used, the sum = (10*2) + (10*5) = 70.
The next smallest possible sums are 70-2 = 68 and 70-2-2 = 66.
Thus, 67 and 69 are not achievable.

Since 4 of 70 possible sums -- 1, 3, 67, and 69 -- are not achievable, the total number of sums that can be formed = 70-4 = 66.

The correct answer is A.

GMATGuruNY wrote:

Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?

A. 66
B. 67
C. 68
D. 69
E. 70

From 1 to 70, there are 70 possible sums.
The smallest answer choice is 66.
Implication:
At most, 4 of the 70 possible sums cannot be made from the given coins.
Impossible sums are likely to be VERY SMALL or VERY LARGE.

Small sums:
Of the sums between 1 and 10, the following cannot be formed from 2¢ and 5¢ coins:
1 and 3.

Large sums:
If all 20 coins are used, the sum = (10*2) + (10*5) = 70.
The next smallest possible sums are 70-2 = 68 and 70-2-2 = 66.
Thus, 67 and 69 are not achievable.

Since 4 of 70 possible sums -- 1, 3, 67, and 69 -- are not achievable, the total number of sums that can be formed = 70-4 = 66.

The correct answer is A.

Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika

mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika

Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.

GMATGuruNY wrote:

mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika

Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.

Whoa!! Thanks so much Mitch!!

Best,
Mallika

GMATGuruNY wrote:

mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika

Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.

Sorry to bug you on this Mitch, but-

67=5+62=5+(2x31)
69=5+64=5+(2x32)

Aren't the above possible..?

Thanks,
Mallika

mallika hunsur wrote:

GMATGuruNY wrote:

mallika hunsur wrote:Hi Mitch,

Does this mean that other sums, can be formed using 5+2c coins and then multiplying by a number..?
Eg: 63= 9x(2+5)..?


But what about other sums like 13, 23, 43 etc..?

Regards,
Mallika

Using up to ten 2¢ coins and ten 5¢ coins, we can form every sum between between 1¢ and 70¢, inclusive, aside from 1, 3, 67 and 69.
We do not need to use the same number of each type of coin.

13¢ can be formed using four 2¢ coins and one 5¢ coin:
(4*2) + 5 = 13.

23¢ can be formed using four 2¢ coins and three 5¢ coins:
(4*2) + (3*5) = 23.

43¢ can be formed using four 2¢ coins and seven 5¢ coins:
(4*2) + (7*5) = 43.

63¢ can be formed using nine 2¢ coins and nine 5¢ coins:
(9*2) + (9*5) = 63.

Sorry to bug you on this Mitch, but-

67=5+62=5+(2x31)
69=5+64=5+(2x32)

Aren't the above possible..?

Thanks,
Mallika

Upto 10 coins, I get it!

mallika hunsur wrote: Sorry to bug you on this Mitch, but-

67=5+62=5+(2x31)
69=5+64=5+(2x32)

Aren't the above possible..?

Thanks,
Mallika

The products in red are not possible because Matt has ONLY 10 of each type of coin.
Thus, 31*2 (implying 31 2¢ coins) and 32*2 (implying 32 2¢ coins) are not viable options.