Place Value and the Number System: Practice

"How many hundreds" asks how many groups of $100$ fit inside the number, not which digit sits in the hundreds column.

$52{,}000 = 520 \times 100$

So $52{,}000$ is $520$ hundreds, even though its hundreds digit is $0$.

Drop each digit into its named column: thousands $7$, hundreds $0$, tens $5$, ones $2$.

$(7 \times 1{,}000) + (0 \times 100) + (5 \times 10) + (2 \times 1) = 7{,}052$

The leftmost $5$ is in the millions place, so its value is $5{,}000{,}000$. The rightmost $5$ is in the ones place, so its value is $5$. Divide the two values.

$\frac{5{,}000{,}000}{5} = 1{,}000{,}000$

The answer is the ratio of the two place values, because the same digit $5$ sits in both columns.

Both numbers share $5$, $3$ in the thousands and hundreds places, so the tens place decides it: the blank must be less than $6$.

$\square < 6 \;\Rightarrow\; \square = 5 \text{ at most}$

With $5$ in the tens place, $5{,}351 < 5{,}361$. Any larger digit makes it $\ge 5{,}361$.

A four-digit number cannot start with $0$, so use the smallest nonzero digit, $1$, in the thousands place. Then fill the remaining places in increasing order with $0, 4, 7$.

$1 \mid 0 \mid 4 \mid 7 = 1{,}047$

The hundreds digit is $3$. Increasing only that digit by $4$ changes it to $7$, and every other place stays put.

$3 + 4 = 7 \;\Rightarrow\; 6{,}352 \to 6{,}752$

Group into periods from the right: $6$ millions, $470$ thousands, $092$ ones. The $4$ occupies the hundreds slot of the thousands period, which is the hundred-thousands place.

$4 \times 100{,}000 = 400{,}000$

Write one three-digit block for each period: $12$ millions, $004$ thousands, $007$ ones. Every period after the leading one is padded out to three columns with zeros.

$12{,}004{,}007$

Drop one of those padding zeros and the later digits land in the wrong columns.

All four share $8$ thousands, so compare the hundreds place next. Two have $0$ hundreds ($8{,}019$ and $8{,}091$) and two have $1$ hundred ($8{,}109$ and $8{,}190$), so the $0$-hundreds pair comes first; break each tie on the tens digit.

$8{,}019 < 8{,}091 < 8{,}109 < 8{,}190$

A value of $30{,}000$ means the $3$ sits in the ten-thousands place.

$3 \times 10{,}000 = 30{,}000$

In $430{,}000$ the $3$ is in the ten-thousands place. (In the others the $3$ is in the millions, thousands, or hundred-thousands place.)

Between $200$ and $300$ forces a $2$ in the hundreds place, and the ones digit is $0$, so the number looks like $2\square 0$. The digit sum must be $6$, so the tens digit is $6 - 2 - 0 = 4$.

$2 + 4 + 0 = 6 \;\Rightarrow\; 240$

Only $240$ meets all three conditions.

Compare each candidate to the two bounds, place by place from the left.

$6{,}300 < 6{,}372 < 6{,}400$

The others fall outside. $6{,}289$ drops below the lower bound in the hundreds place, since $2 < 3$. Against $6{,}400$, the numeral $6{,}403$ first differs in the ones place, and $6{,}430$ first differs in the tens place, so both are larger than the upper bound.