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Private: Learning Math: Number and Operations
Divisibility Tests and Factors Part A: Alpha Math (35 minutes)
“Alpha math” problems, where each letter stands for one digit of a number, can help you identify some of the things you know about the behaviors of particular base ten digits under various operations. Your task is to decode each of the following problems, figuring out what digit each letter represents. See Note 1 below.
Problem A1
In the following sums, one letter always represents the same digit in each problem, and no digit is represented by more than one letter. Replace the letters with digits:
Consider the following:
• What does the sum of a and a in the hundreds column tell you about the value of a?
• Notice that the sum of b and c in the tens column is different from the sum of c and b in the ones column. What does this tell you about the value of b + c?
• Notice that the sum of b and c in the tens column is b. What does this tell you about c?
Problem A2
Each letter represents a different digit of a number:
Decode the problem to determine the following value:
Problem A3
In these problems, the asterisks represent missing digits (though they do not all represent the same digit, as do the letters in the previous problems). Identify the missing digits in the following multiplication problems:
Problem A4
Each letter represents a different digit of a three-digit number. Decode the problem:
TAKE IT FURTHER
Problem A5
Jen has found a special five-digit number that she calls abcde. If you enter the number 1 and then her number on a calculator and multiply it by 3, the result is the same number with a 1 on the other end:
1abcde • 3 = abcde1
What is her number?
Notes
Note 1
Doing Alpha math is like decoding a cipher — it helps the solver think about how different number/letter combinations relate to one another. For example, ab + b = cdd suggests that a must be 9, because no other digit in the tens place would give a three-digit sum. The most that could be added to 9 in the tens column is 1, because two one-digit numbers cannot add to more than 18. That means that d must be 0. Since the sum has equal ones and tens digits, b must be 5. This type of reasoning is an important step toward a deep understanding of the operations involved in the Alpha math problems.
Solutions
Problem A1
a. The number y must be either 5 or 0. Since the sum is not 0, y is 5 and m is 1.
b. The digits m and a must be consecutive digits, since (in the tens place) m plus carry equals a. So a is 9 and m is 8.
c. Since the sum is two digits, the digit l must be 1 or 2. It cannot be 1, since the sum of four identical numbers is always even. So l is 2, and n must be 3 (since 3 • 4 = 12), and g is 9.
d. Since x plus carry equals ba, x must be 9 and ba must be 10. So x is 9, b is 1, and a is 0 (the sum is 999 + 1 = 1,000).
Problem A2
The number a2 is a two-digit number, with digits different from a. The number a3 is a three-digit number, with different digits from both a and a2. Here are the possibilities:
a = 5; a2 = 25; a3 = 125 (no, since c = a = 5)
a = 6; a2 = 36; a3 = 216 (no, since c = a = 6)
a = 7; a2 = 49; a3 = 343 (no, since d = f = 3)
a = 8; a2 = 64; a3 = 512 (yes!)
a = 9; a2 = 81; a3 = 729 (no, since a = f = 9)
The value of bc – a = 64 – 8 = 56.
Problem A3
a. The solution is 169 • 7 = 1,183. It is easiest to first find the upper-right asterisk, then use the known carry digits to fill in the rest of the product.
b. The solution is 47 • 9 = 423. The units-digit asterisk must be 9, since 63 is the only multiple of 7 that ends in 3. Then, with the carry digit 6, only 4 • 9 = 36 + 6 = 42 can give the leading digit of 4.
c. The solution is 64 • 6 = 384. The digit being multiplied by must be either 5 or 6 (to give a hundreds digit of 3); if it is 5, the units-digit would have to be 0 or 5. Since it is 4, the digit being multiplied by must be 6. Then the remaining units-digit asterisk must be either 9 or 4; 69 • 6 = 414 is not valid, so the remaining asterisk is 4.
Problem A4
The fact that the sum of a and a equals c, a single digit, means that a can be no more than 4. The fact that the sum of c and b is two different values means that b + c must be larger than 10; it also means that a and b are consecutive numbers. Since the sum of b + c (plus any carry digit) equals b, then c must be either 0 or 9. Knowing that b + c is greater than 10 means that c must equal 9. So the sum is now:
Finding a is next. Since a + a (plus any carry digit) equals 9, a must be 4. Then b must be 5, since 9 + b = a, with no carry possibility. The final sum is:
Problem A5
We can solve this by building abcde starting with e. Since e • 3 ends in 1, e must be 7. Then d • 3 + 2 (carry) = 7, so d is 5. Then c • 3 + 1 (carry) = 5, so c is 8. Then b • 3 + 2 (carry) = 8, so b is 2. Then a • 3 + 0 (no carry) = 2, so a is 4.
The number abcde is 42,857, and the equation is 142,857 • 3 = 428,571.
