Lorna collected 1/3 the number of coins that Mandy collected. Mandy collected two and one quarter (2 1/4) the number of coins collected by Enver. If all three collected 480 coins then how many coins did Mandy collect?

Let $M$ be the number of coins Mandy collected.
Let $L$ be the number of coins Lorna collected.
Let $E$ be the number of coins Enver collected.
From the problem:

1. $L = \frac{1}{3} M$
2. $M = 2\frac{1}{4} E = \frac{9}{4} E$
3. $L + M + E = 480$

From (1), $L = \frac{1}{3} M$.
From (2), we can express $E$ in terms of $M$: $E = M \div \frac{9}{4} = M \times \frac{4}{9} = \frac{4}{9} M$.

Find a common denominator, which is 9.

Mandy collected 270 coins.

The final answer is $\boxed{\text{270 coins}}$.

• Numbers with '4' in the units place: 84, 94, 104, 114, 124, 134, 144, 154, 164, 174. (10 occurrences)
• Numbers with '4' in the tens place: 140, 141, 142, 143, 144, 145, 146, 147, 148, 149. (10 occurrences) The number 144 contains two '4's (one in the units place and one in the tens place). Total '4's = 10 (from units place) + 10 (from tens place) = 20.

• Numbers with '8' in the units place: 88, 98, 108, 118, 128, 138, 148, 158, 168, 178. (10 occurrences)
• Numbers with '8' in the tens place: 80, 81, 82, 83, 84, 85, 86, 87, 88, 89. (10 occurrences) The number 88 contains two '8's (one in the units place and one in the tens place). Total '8's = 10 (from units place) + 10 (from tens place) = 20. Wait, I need to be careful with the range. For 8's in the tens place, the numbers are 80-89. For 8's in the units place, the numbers are 88, 98, 108, 118, 128, 138, 148, 158, 168, 178. Let's list them and count: For '8's: 80, 81, 82, 83, 84, 85, 86, 87, 88 (two 8s), 89 (10 numbers, 11 '8's) 98 (1 '8') 108 (1 '8') 118 (1 '8') 128 (1 '8') 138 (1 '8') 148 (1 '8') 158 (1 '8') 168 (1 '8') 178 (1 '8') Total '8's = 11 + 9 = 20.

Let's re-verify the 8's count.
Numbers from 80 to 180.
Units digit is 8: 88, 98, 108, 118, 128, 138, 148, 158, 168, 178. (10 numbers, 10 '8's from units place)
Tens digit is 8: 80, 81, 82, 83, 84, 85, 86, 87, 88, 89. (10 numbers, 10 '8's from tens place)
The number 88 has two '8's. So, the total count of '8's is 10 + 10 = 20.

My previous manual count was 21. Let's re-do it carefully.
80 (1)
81 (1)
82 (1)
83 (1)
84 (1)
85 (1)
86 (1)
87 (1)
88 (2)
89 (1)
Subtotal for 80-89: 1+1+1+1+1+1+1+1+2+1 = 11 '8's.

90-99: 98 (1 '8')
100-109: 108 (1 '8')
110-119: 118 (1 '8')
120-129: 128 (1 '8')
130-139: 138 (1 '8')
140-149: 148 (1 '8')
150-159: 158 (1 '8')
160-169: 168 (1 '8')
170-179: 178 (1 '8')
180: No '8'
Subtotal for 90-180: 1+1+1+1+1+1+1+1+1 = 9 '8's.

Total '8's = 11 + 9 = 20.
So, there are 20 '4's and 20 '8's.

Total = (Number of '4's) + (Number of '8's)
Total = 20 + 20 = 40.

The final answer is $\boxed{\text{40}}$.

The pattern consists of blocks, each starting with two white stones (WW) followed by a number of blue stones (B).
Block 1: W W B B (2 White, 2 Blue) - Total 4 stones
Block 2: W W B B B B (2 White, 4 Blue) - Total 6 stones
Block 3: W W B B B B B B B B (2 White, 8 Blue) - Total 10 stones
It appears the number of blue stones in each block is a power of 2: $2^1, 2^2, 2^3, \dots, 2^n$.
So, Block $n$ has 2 White stones and $2^n$ Blue stones, for a total of $2 + 2^n$ stones.

• Block 1 ($n=1$): 2W, $2^1$B = 2B. Total stones = 4. Cumulative stones = 4. Cumulative blue = 2.
• Block 2 ($n=2$): 2W, $2^2$B = 4B. Total stones = 6. Cumulative stones = 4 + 6 = 10. Cumulative blue = 2 + 4 = 6.
• Block 3 ($n=3$): 2W, $2^3$B = 8B. Total stones = 10. Cumulative stones = 10 + 10 = 20. Cumulative blue = 6 + 8 = 14.
• Block 4 ($n=4$): 2W, $2^4$B = 16B. Total stones = 18. Cumulative stones = 20 + 18 = 38. Cumulative blue = 14 + 16 = 30.
• Block 5 ($n=5$): 2W, $2^5$B = 32B. Total stones = 34. Cumulative stones = 38 + 34 = 72. Cumulative blue = 30 + 32 = 62.

The pendant has a total of 108 stones. After 5 full blocks, 72 stones have been used.
Remaining stones = $108 - 72 = 36$ stones.

The next block (Block 6, $n=6$) would start with 2 White stones, followed by $2^6 = 64$ Blue stones.
The 36 remaining stones must come from Block 6.
The first 2 stones of Block 6 are White (WW). So, 2 White stones are added.
The number of stones remaining to be placed after the 2 white stones is $36 - 2 = 34$.
These 34 stones must be blue stones from Block 6. This also satisfies the condition that the last stone is blue.

Total blue stones = (Cumulative blue stones from 5 full blocks) + (Blue stones from the partial 6th block)
Total blue stones = $62 + 34 = 96$.

The final answer is $\boxed{\text{96 blue stones}}$.