Question

problem 20

\

$$\begin{array}{r} \\text{she} \\\\ \\times \\quad \\text{he} \\\\ \\hline \\text{here} \\\\ \\text{she} \\\\ \\hline \\text{true} \\end{array}$$

\text{e} = ? \quad \text{h} = ? \quad \text{r} = ? \quad \text{s} = ? \quad \text{t} = ? \quad \text{u} = ?

clue: notice that e times e equals a number that ends in e. what three numbers might e equal? which one works?

Explanation:

Analyze the cryptarithm structure

We are given the multiplication puzzle:

$$
LATEXBLOCK0
$$

This represents the standard multiplication algorithm:

1. \(\text{she} \times e = \text{here}\)
2. \(\text{she} \times h = \text{she}\) (shifted by one position to the left)
3. \(\text{here} + 10 \times \text{she} = \text{true}\)

Let's represent the letters as distinct digits from \(0\) to \(9\): \(s, h, e, r, t, u\).

Determine the value of \(h\)

From the second partial product:

$$ \text{she} \times h = \text{she} $$

Since \(\text{she}\) is a non-zero three-digit number, multiplying it by \(h\) yields itself.
Therefore, we must have:

$$ h = 1 $$

Analyze the first partial product

Using \(h = 1\), the first partial product is:

$$ \text{s1e} \times e = \text{1ere} $$

Let's look at the units digit of this multiplication:

$$ e \times e \equiv e \pmod{10} $$

The single digits that satisfy \(e^2 \equiv e \pmod{10}\) are \(e \in \{0, 1, 5, 6\}\).

• Since \(h = 1\) and digits must be distinct, \(e

eq 1\).

• If \(e = 0\), then \(\text{s10} \times 0 = 0\), which does not equal the four-digit number \(\text{10r0}\). Thus, \(e

eq 0\).
Therefore, \(e\) must be either \(5\) or \(6\).

Test the case \(e = 5\)

If \(e = 5\), the multiplication is:

$$ \text{s15} \times 5 = \text{15r5} $$

Let's expand this algebraically:

$$ (100s + 15) \times 5 = 1505 + 10r $$

$$ 500s + 75 = 1505 + 10r $$

$$ 500s - 10r = 1430 $$

Divide by 10:

$$ 50s - r = 143 $$

Since \(r\) is a single digit (\(0 \le r \le 9\)):

• If \(s = 3\): \(150 - r = 143 \implies r = 7\).

Let's check if this assignment works:
Digits used so far: \(h=1\), \(e=5\), \(s=3\), \(r=7\). These are all distinct.
Let's calculate the final sum \(\text{true}\):

$$ \text{true} = \text{here} + 10 \times \text{she} = 1575 + 3150 = 4725 $$

Matching \(\text{true} = 4725\):

• \(t = 4\)
• \(r = 7\) (consistent with above)
• \(u = 2\)
• \(e = 5\) (consistent with above)

Let's check all digits for uniqueness:

• \(s = 3\)
• \(h = 1\)
• \(e = 5\)
• \(r = 7\)
• \(t = 4\)
• \(u = 2\)

All digits \(\{1, 2, 3, 4, 5, 7\}\) are distinct and valid.

Test the case \(e = 6\)

If \(e = 6\), the multiplication is:

$$ \text{s16} \times 6 = \text{16r6} $$

$$ (100s + 16) \times 6 = 1606 + 10r $$

$$ 600s + 96 = 1606 + 10r $$

$$ 600s - 10r = 1510 $$

Divide by 10:

$$ 60s - r = 151 $$

• If \(s = 3\): \(180 - r = 151 \implies r = 29\) (not a single digit).
• If \(s = 2\): \(120 - r = 151 \implies r = -31\) (not a single digit).

Thus, no solution exists for \(e = 6\).

Answer:

Question 1

she \(\times\) he = true:
\(e = 5\), \(h = 1\), \(r = 7\), \(s = 3\), \(t = 4\), \(u = 2\)

\(e = \) <blank>5</blank> \(h = \) <blank>1</blank> \(r = \) <blank>7</blank> \(s = \) <blank>3</blank> \(t = \) <blank>4</blank> \(u = \) <blank>2</blank>