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[x] is the greatest integer less than or equal to x. Find the number Updated on: 30 Jun 2019, 22:14
Originally posted by CareerGeek on 30 Jun 2019, 01:47.
Last edited by CareerGeek on 30 Jun 2019, 22:14, edited 1 time in total.
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95% (hard)

Question Stats:

43% (02:26) correct 57% (02:17) wrong based on 107 sessions

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[x] is the greatest integer less than or equal to x. Find the number of positive integers n such that [\(\frac{n}{7}\)] = [\(\frac{n}{9}\)]

A. 12
B. 13
C. 14
D. 15
E. 16
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D
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[x] is the greatest integer less than or equal to x. Find the number 30 Jun 2019, 03:02
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When 1≤n≤6, [n/7]=[n/9]=0
When 9≤n≤13, [n/7]=[n/9]=1
When 18≤n≤20, [n/7]=[n/9]=2
When n=27, [n/7]=[n/9]=3

Total possible value of n= 6+5+3+1=15

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[x] is the greatest integer less than or equal to x. Find the number of positive integers n such that [\(\frac{n}{7}\)] = [\(\frac{n}{9}\)]

A. 12
B. 13
C. 14
D. 15
E. 16
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[x] is the greatest integer less than or equal to x. Find the number 01 Jul 2019, 09:46
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Hi, can someone explain the problem?
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[x] is the greatest integer less than or equal to x. Find the number 02 Jul 2019, 18:48
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Dillesh4096
[x] is the greatest integer less than or equal to x. Find the number of positive integers n such that [\(\frac{n}{7}\)] = [\(\frac{n}{9}\)]

A. 12
B. 13
C. 14
D. 15
E. 16
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We see that when n is any integer from 1 to 6, inclusive, then [n/7] = 0 = [n/9]. Similarly, when n is any integer from 9 to 13, inclusive, then [n/7] = 1 = [n/9]; when n is any integer from 18 to 20, inclusive, then [n/7] = 2 = [n/9]; and lastly, when n = 27, [n/7] = 3 = [n/9].

Therefore, there are 6 + 5 + 3 + 1 = 15 integer values for n.

Answer: D
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[x] is the greatest integer less than or equal to x. Find the number 25 Nov 2025, 03:37
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abhishek31
Hi, can someone explain the problem?
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Let \([x]\) denote the greatest integer less than or equal to \(x\). We are looking for the number of positive integers \(n\) such that:
\([\frac{n}{7}] = [\frac{n}{9}]\)

Let's call this common integer value \(k\).
By definition of the floor function, this implies two inequalities:
1. \(k \le \frac{n}{7} < k + 1 \implies 7k \le n < 7k + 7\)
2. \(k \le \frac{n}{9} < k + 1 \implies 9k \le n < 9k + 9\)

Method: Intersection of Intervals
For a valid \(n\) to exist for a specific \(k\), there must be an overlap between the interval \([7k, 7k+7)\) and the interval \([9k, 9k+9)\).
The overlap exists only if the starting point of the second interval is less than the ending point of the first interval.

CRITICAL CHECK: Find the maximum possible value for \(k\).
\(9k < 7k + 7\)
\(2k < 7\)
\(k < 3.5\)
Since \(k\) must be an integer, the possible values for \(k\) are 0, 1, 2, and 3.

Now, let's count the valid integers \(n\) for each case. Remember that \(n\) must be a positive integer.

***

Case 1: \(k = 0\)
Range 1: \(0 \le n < 7\)
Range 2: \(0 \le n < 9\)
Intersection: \(0 \le n < 7\)
Possible integers: \(0, 1, 2, 3, 4, 5, 6\)
Since \(n\) must be positive, we exclude 0.
Valid values: \(\{1, 2, 3, 4, 5, 6\}\) -> 6 values

Case 2: \(k = 1\)
Range 1: \(7 \le n < 14\)
Range 2: \(9 \le n < 18\)
Intersection: \(9 \le n < 14\)
Valid values: \(\{9, 10, 11, 12, 13\}\) -> 5 values

Case 3: \(k = 2\)
Range 1: \(14 \le n < 21\)
Range 2: \(18 \le n < 27\)
Intersection: \(18 \le n < 21\)
Valid values: \(\{18, 19, 20\}\) -> 3 values

Case 4: \(k = 3\)
Range 1: \(21 \le n < 28\)
Range 2: \(27 \le n < 36\)
Intersection: \(27 \le n < 28\)
Valid values: \(\{27\}\) -> 1 value

***

Total Count
Summing up the values from all cases:
\(Total = 6 + 5 + 3 + 1 = 15\)

This matches Option (D).

Answer: D
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[x] is the greatest integer less than or equal to x. Find the number 03 Dec 2025, 23:56
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So.... putting any number inside [...] means it will show an integer....but if da number is fraction...then da fraction part will be removed and rounded off....example [ 9.98 ] = 9....and [ 9.13 ] = 9......
Now....we gotta make [ n / 7 ] = [ n / 9 ] ......

If...n = 1, 2, 3, 4, 5, 6.....then...... 0 < [ n / 7 ] < 1 .... so [ n / 7 ] = 0
..... And..... 0 < [ n / 9 ] < 1 .... so [ n / 9 ] = 0
So ..... [ n / 7 ] = [ n / 9 ] holds.......

if n = 7,8 ....then 1 ≤ [ n / 7 ] < 2 ... so [ n / 7 ] = 1
..... And..... 0 < [ n / 9 ] < 1 .... so [ n / 9 ] = 0
So.... [ n / 7 ] ≠ [ n / 9 ]

If n = 9, 10, 11, 12, 13....then.... 1 < [ n / 7 ] < 2 ... So [ n / 7 ] = 1
..... And..... 1 ≤ [ n / 9 ] < 2 .... so [ n / 9 ] = 1
So ..... [ n / 7 ] = [ n / 9 ] holds.......

If n = 14, 15, 16, 17 ....then.... 2 ≤ [ n / 7 ] < 3 ... So [ n / 7 ] = 2
.... And..... 1 < [ n / 9 ] < 2 .... so [ n / 9 ] = 1
So.... [ n / 7 ] ≠ [ n / 9 ]

If n = 18, 19, 20 .....then.... 2 < [ n / 7 ] < 3 ... So [ n / 7 ] = 2
.... And..... 2 ≤ [ n / 9 ] < 3 .... so [ n / 9 ] = 2
So ..... [ n / 7 ] = [ n / 9 ] holds.......

If n = 21, 22, 23, 24, 25, 26 .....then.... 3 ≤ [ n / 7 ] < 4 ... So [ n / 7 ] = 3
.... And..... 2 < [ n / 9 ] < 3 .... so [ n / 9 ] = 2
So.... [ n / 7 ] ≠ [ n / 9 ]

If n = 27 ....then.... 3 < [ n / 7 ] < 4 ... So [ n / 7 ] = 3
.... And..... [ n / 9 ] = 3 .... so [ n / 9 ] = 3
So ..... [ n / 7 ] = [ n / 9 ] holds.......

But for n = 28, 29, 30, 31, 32, 32, 34 .... 4 ≤ [ n / 7 ] < 5 ..... But 3 < [ n / 9 ] < 4 .... So.... [ n / 7 ] ≠ [ n / 9 ] .....
But anymore than n = 34 and something aberrant happens....
.... If n = 35 ..... [ n / 7 ] = 5 but.... 3 < [ n / 9 ] < 4 ...so... [ n / 9 ] = 3
Now... [ n / 7 ] becomes 5 but... [ n / 9 ] stays 3 .... So difference is 2 ... Up until now difference had been 1 .....so it means from now.... [ n / 9 ] will never catch up to [ n / 7 ] again......

So all possible values of ... n ... is 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 13, 18, 19, 20 and 27 .....
So 15 possible n ......
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